Latin America Resources Essay -- essays research papers

I. 1.Miguel Angel Asturias- A Nobel Prize wining novelist and poet from Guatemala. Wrote about his experience under a dictatorship. Established small-town life and a clash of cultures as themes 2.Frida Kahlo- Started painting in 1925 when she was hospitalized. Married Diego Rivera. Was inspired by retablos, religious paintings. Was a champion of Mexican culture. 3.John F. Kennedy – Kennedy encouraged Latin American countries to undertake reforms to raise the standard of living for their people with the Alliance for Progress in 1961. 4.Luis Munoz Marin – Became Puerto Rico’s first elected governor in the 1950s. He supported a program to encourage tourism and develop industry on the island. 5.Gabriel Garcia Marquez- Nobel Prize winning author in Columbia. Wrote in a Style called magical realism. Most popular novel was One Hundred years of Solitude. 6.James Munroe – Issued the Munroe Doctrine in 1823 which disallowed colonization in the Americas 7.Pablo Neruda - Chile’s Nobel Prize winning poet who criticized the United States for using its power and wealth to carve up Panama. 8.Manuel Noriega – Panama’s president who was charged by the United States with drug trafficking in 1988. 9.Franklin D. Roosevelt – Announced the Good Neighbor Policy in 1933, which declared that â€Å"no state has the right to intervene in the internal or external affairs of another state.† 10.Theodore Roosevelt – President that offered Columbia $10 million for a strip of land in Panama to build a canal. Roosevelt encouraged rebels in Panama to rebel when Columbia rejected the offer. In 1903 when the Panama people received their independence, they granted US the 10 mile wide â€Å"canal zone.† II. 1. Under the Platt Amendment, the United States claimed the right to intervene in Cuban affairs. 2. The United States gained Puerto Rico and Philippines from Spain. 3. Financial interests led the United States to intervene in Latin America. In the early 1900s, the Dominican Republic was unable to pay its debts to American banks. American forces also occupied Nicaragua and Haiti ,and intervened in the affairs of Honduras 6 times. In each case, they stepped in to protect American lives and property or to support a government that favored American interests. 4. The United States wanted to create a solid anti-communist bloc in the Americas. The United States’ primary concern was the communist country of ... ...lopment and democracy. The OAS includes the United States, which has tended to dominated the organization. 11.retablos – religious paintings offered as thanks for escape from misfortune. 12.Sendero Luminoso – (Shining Path) Extreme left wing group that resorted to torture and murder in their on-going struggle to control the country of Peru. 13.Spanish-American War – As the United States industrialized, it extended its influence in the Caribbean and Central America. During the 1890s, Cuban patriots were battling for independence from Spain. In 1898, the United States declared war on Spain and joined the fighting. It promised that once peace was achieved it would â€Å"leave the government and control of Cuba to its people.† The Spanish American War ended in victory for the United States. Cuba did gain its independence, but the United States forced Cuba to include the Platt Amendment as part of its new constitution. Under the Platt Amendment the United States claimed the right to intervene in Cuban affairs. As a result the United States gained Puerto Rico and Philippines from Spain. 14.ultraismo – The belief that art should exist for its own sake, not for any social or political reasons.

NHL Players Moving West :: essays research papers

  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   NHL Players Moving East   Ã‚  Ã‚  Ã‚  Ã‚  It was deemed official this past weekend that the NHL will be on lockout due to the new salary cap the league wants to enforce. The players in the league have a different outlook this year as well and it involves moving east. Hockey players in the NHL get 75% of the revenue the league makes and it is very obvious that this current situation is not going to last. There is no way to fund the league if the players are making all of the money. So what happens now? Ratings are very low, the league is not m   Ã‚  Ã‚  Ã‚  Ã‚  September 21, 2004   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Current Events Paper arketable right now, and expansion has hurt rather than cured many of the problems in the NHL. That is why the NHL is trying to come to an agreement to place a cap in the league in order to make some much-needed revenue. The players are not having this new idea, and they are now welcoming a different home.   Ã‚  Ã‚  Ã‚  Ã‚  More than 150 players have signed to play in the European leagues. Jaromir Jagr has agreed to play in the Czech Republic for a team named Kladmo. Marcus Naslund has agreed to play in Sweden for Modo. Llya Kovalchuk has signed with AK Bars Kazan in Russia. These are just a few names, but other players are already signed as well and are playing games as we speak. The Russian league has signed 33 NHL players, the Swedish league has signed 30 NHL players, the Czech league has signed 47 NHL players, and the Finnish and Slovakia leagues both signed nine NHL players apiece.   Ã‚  Ã‚  Ã‚  Ã‚  So how can this current situation be resolved? There are few options right now, which makes it seem like this is really going to hurt the league economically. Most of the players in the league have a lockout clause in their contracts making this problem easy for them to deal with. All they have to do is wait out this lockout while being involved with a different league and when the lockout is over they can automatically resume their previous positions in the NHL. This situation however, is not going to be resolved that easily. If the players do not agree to have a salary cap in the NHL then there is not quite an alternative. Where can the league make up this money? If there were a bigger demand for the sport than there would not be a problem.

Nss Phy Book 2 Answer

1 1 2 3 C Motion I 7 (a) From 1 January 2009 to 10 January 2009, the watch runs slower than the actual time by 9 minutes. Therefore, when the actual time is 2:00 pm on 10 January 2009, the time shown on the watch should be 1:51 pm on 10 January 2009. Practice 1. 1 (p. 6) D (a) Possible percentage error 10 ? 6 = ? 100% 24 ? 3600 = 1. 16 ? 10 % 1 (b) = 1 000 000 days 10 ? 6 –9 It would take 1 000 000 days to be in error by 1 s. (b) Percentage error 9 = ? 100% 9 ? 24 ? 60 = 6. 94 ? 10–2% 4 (a) One day = 24 ? 60 ? 60 = 86 400 s Practice 1. 2 (p. 15) 1 2 3 4 5 C B D D (b) One year = 365 ? 86 400 = 31 500 000 s 5 Let t be the period of time recorded by a stop-watch. Percentage error = 0. 4 ? 100% ? 1% t t ? 40 s (a) Total distance she travels 2 ? ? 10 2 ? ? 20 2 ? ? 15 + + = 2 2 2 = 141 m (b) Magnitude of total displacement = 10 ? 2 + 20 ? 2 + 15 ? 2 = 90 m Direction: east Her total displacement is 90 m east. The minimum period of time is 40 s. 6 (a) Percentage error error due to reaction time = ? 100% time measured 0. 3 = ? 100% 10 = 3% 6 7 His total displacement is 0. With the notation in the figure below. (b) From (a), the percentage error of a short time interval (e. g. 10 s) measured by a stop-watch is very large. Since the time intervals of 110-m hurdles are very short in the Olympic Games, stop-watches are not used to avoid large percentage errors. Since ZX = ZY = 1 m, ? = ? = 60 °. Therefore, XY = ZX = ZY = 1 m The magnitude of the displacement of the ball is 1 m.  © 8 (a) The distance travelled by the ball will be longer if it takes a curved path. 7 (a) Length of the path = 0. 8 ? 120 = 96 m (b) No matter which path the ball takes, its displacement remains the same. (b) Length of AB along the dotted line 96 = 30. 6 m = (c) Magnitude of Jack’s average velocity 30. 6 ? 2 = = 0. 51 m s–1 120 Practice 1. 3 (p. 23) 1 B Total time 5000 5000 = + = 9821 s 1. 4 0. 8 5000 + 5000 = 1. 02 m s–1 Average speed = 9821 Practice 1. 4 (p. 31) 1 2 C B Final speed = 1. 5 ? 1 – 0. 2 ? 1 = 1. 3 m s–1 2 C Total time = 9821 + 10 ? 60 =10 421 s 5000 + 5000 Average speed = = 0. 96 m s–1 10 421 3 A By a = 3 D When the spacecraft had just finished 1 revolution, the spacecraft returned to its starting point. Therefore, its displacement was zero and its average velocity was also zero. v ? u , t v = u + at 36 = + ( ? 1. 5) ? 2 3. 6 = 7 m s–1 = 7 ? 3. 6 km h–1 = 25. 2 km h–1 Its speed after 2 s is 25. 2 km h–1. 4 5 D (a) Average speed 100 = = 10. m s–1 9. 69 (b) Yes. This is because the magnitude of the displacement is equal to the distance in this case. 4 B Take the direction of the original path as positive. Average acceleration of the ball ? 10 ? 17 = 0. 8 = –33. 8 m s–2 The magnitude of the average acceleration of the ball is 33. 8 m s– 2. v ? u By a = , t 100 ? 0 v ? u 3. 6 t= = = 4. 27 s a 6. 5 6 (a) Two cars move with the same speed, e. g. 50 km h–1, but in opposite directions. (b) A man runs around a 400-m playground. When we calculate his average speed, we can take 400 m as the distance and his average speed is non-zero. But since his displacement is zero (he returns to his starting point), his average velocity is zero. 5 The shortest time it takes is 4. 27 s.  © 6 Time / s –1 4 0 2 4 6 17 8 22 D Average speed 80 + 60 = 5 = 28 km h–1 Average velocity = Speed / m s 2 7 12 v ? u 22 ? 2 a= = 2. 5 m s–2 = t 8 The acceleration of the car is 2. 5 m s–2. 7 (a) I will choose ‘towards the left’ as the positive direction. 80 2 + 60 2 5 (b) 5 = 20 km h–1 C Total time 10 10 = + 2 3 = 8. 33 s v ? u , t u = v ? at = 9 ? (? 2) ? 3 = 15 m s–1 –1 (c) By a = Average speed 20 = 8. 33 = 2. 4 m s–1 Her average speed for the whole trip is 2. m s–1. The initial velocity of the skater is 15 m s . 8 (a) The object initially moves towards the left and accelerates towards the left. It will speed up. 6 7 8 9 10 C C C B A Magnitude of displacement = 2000 2 + 6000 2 = 6324. 6 m Magnitude of average velocity 6324. 6 = 4 ? 3600 = 0. 439 m s–1 6000 tan ? = 2000 ? = 71. 6 ° His average velocity is 0. 439 m s–1 (S 71. 6 ° E). (b) The object initially moves towards the right and accelerates towards the left. It will slow down. Its velocity will be zero and then increases in the negative direction (moves towards the left). Revision exercise 1 Multiple-choice (p. 5) 1 2 3 C D B  © 11 C Total time = 13 min = 780 s 840 ? 2 = 2. 15 m s ? 1 Average speed = 780 (b) Displacement from Sheung Shui to Lok Ma Chau 1000 = ? 6. 3 1 = 6300 m Magnitude of average velocity 6300 = 359 = 17. 5 m s–1 (1M) (1A) (1M) (1A) 12 13 D (HKCEE 2003 Paper II Q3) Conventional (p. 37) 1 Total time left for the two players = 4 ? 60 + 9 + 5 ? 60 + 16 = 565 s Total time they have been playing = 2 ? 60 ? 60 ? 565 = 6635 s (= 110 min 35 s = 1 h 50 min 35 s) (1A) 5 (a) Total distance = 1500 + 40 ? 1000 + 10 ? 1000 = 51 500 m Total time = 2 ? 3600 + 3 ? 60 + 8 = 7388 s Average speed 51 500 = 7388 = 6. 7 m s–1 (1M) (1A) 2 (a) 50 m (1A) (b) Ma gnitude of average velocity of Kitty 50 = (1M) 1? 60 + 15 = 0. 667 m s ? 1 (1A) (1M) (1A) (c) Average speed of the coach 5 + 50 + 5 = 1? 60 + 15 = 0. 8 m s ? 1 (b) Swimming: Average speed 1500 = 21 ? 60 + 28 = 1. 16 m s–1 Cycling: Average speed 40 000 = 1 ? 3600 + 1 ? 60 + 53 = 10. 8 m s–1 Running: Average speed 10 000 = 39 ? 60 + 47 = 4. 19 m s–1 (1M) His average speed was the highest in cycling. (1A) 3 (a) Since she measures the time interval based on 1 cycle of the pendulum, the error (0. 3 s) in measuring the cycle of the pendulum accumulates. is from 8 to 14 s. 1A) (1A) The range of the time interval (10 cycles) (b) When finding the time for one pendulum cycle, Jenny should time more pendulum cycles (e. g. 20) with the stop-watch and divide the time by the number of cycles. (1A) 4 (a) Time required 7. 4 ? 1000 = 20. 6 = 359 s (5 min 59 s) (1M) (1A)  © (c) Yes. Since the time interval of this competition is quite long, (1A) using stop-watch will not result in large percentage error as the reaction time for an average person is only 0. 2 s. (1A) (1M) (c) Total time = 5 min 45 s ? 1 min 58 s = 3 min 47 s = 3 ? 60 + 47 = 227 s v? u a= (1M) t 431 ? 0 = 3. = 0. 527 m s–2 (1A) 227 The average acceleration of the train is 0. 527 m s–2. 6 (a) v = u + at =0+6? 4 = 24 m s–1 = 86. 4 km h 86. 4 km h . –1 –1 (1A) The maximum speed of the car is 8 (1M) (a) Total distance = 8000 + 4000 + 5000 = 17 000 m Total time = 1 ? 3600 + 30 ? 60 + 45 ? 60 (b) v = u + at = 24 + (–4) ? 2 = 16 m s –1 –1 = 57. 6 km h (1A) –1 = 8100 s Average speed 17 000 = 8100 = 2. 10 m s–1 (1M) (1A) (c) The final speed of the car is 57. 6 km h . v? u a= (1M) t 16 ? 0 = 6 = 2. 67 m s–2 2. 67 m s–2. (1A) The average acceleration of the car is (b) 7 (a) Average speed 30 000 = 8 ? 60 = 62. m s–1 The average speed of the train is 62. 5 m s–1. (1M) (1A) (b) Maximum speed 430 = = 119. 4 m s? 1 > average speed 3. 6 (1A) The average speed must be smaller than the maximum speed because the train needs to speed up from start and slows down to stop during the trip. (1A) Magnitude of displacement = 3000 2 + 4000 2 = 5000 m Magnitude of average velocity 5000 = = 0. 617 m s–1 8100 4000 tan ? = 3000 (1A) ? = 53. 1 ° His average velocity is 0. 617 m s (N 53. 1 ° E).  © –1 (1A) 9 (a) Distance travelled = 10. 5 ? 3 ? 60 = 1890 m (1M) (1A) 10 (a) Total distance = (120 + 50) ? 1000 = 170 000 m (1M) (1A) b) Circumference of the track =2 r = 2 (400) = 2513 m The distance travelled by Marilyn is 3 1890 m which is about of the 4 circumference. (1A) (b) N ?XYZ is a right-angled triangle. Z ? 50 km 30 ° Y 60 ° X ? ? 120 km Magnitude of displacement (from town X to town Z) = 120 000 2 + 50 000 2 = 130 000 m 120 tan ? = 50 ? = 67. 4 ° Magnitude of displacement AB = 400 2 + 400 2 (1A) (1A) ? = 90 ° ? 67. 4 ° = 22. 6 ° ? = 60 ° ? 22. 6 ° = 37. 4 ° The total displacement of the car is 130 000 m (N 37. 4 ° E). = 566 m Magnitude of average velocity 566 = 3 ? 60 = 3. 14 m s 400 tan ? = 400 ? = 45 ° (S 45 ° E). –1 (c) (1A) Total time 170 000 = = 10 200 s 60 3. 6 Magnitude of average velocity 130 000 = 10 200 = 12. 7 m s–1 Its average velocity is 12. 7 m s (N 37. 4 ° E). –1 (1A) (1A) (1M) (1A) Her average velocity is 3. 14 m s–1  © 11 (a) AC = 60 2 + 80 2 = 100 m 80 tan ? = ? = 53. 1 ° 60 (1M) The total displacement of the athlete is 100 m (S53. 1 °W). (1A) 13 (Correct label of velocity with correct direction (towards the left). ) (Correct label of acceleration with correct direction (towards the right). ) (1A) (1A) (a) The coin moves in the following sequence: B A C C A Therefore, it is at A finally. Displacement of the coin = 15 cm (1A) (1M) (1A) (1M) b) Distance travelled by the coin = 15 + 30 + 30 = 75 cm (b) Time / s v / m s–1 0 –6 1 –4 2 –2 3 0 4 +2 5 +4 6 +6 (1A) (1A) (c) (i) Total time = 2 s ? 4 = 8 s Average velocity 15 ? 10 ? 2 = 8 = 0. 0188 m s? 1 (0. 5A ? 6) (1M) (1A) (c) The car will slow down and its speed will drop to zero. After th at the car will move towards the right with increasing speed (uniform acceleration). (1A) (1M) (1A) (1M) (1A) (1M) (1A) A (ii) Average speed 75 ? 10 ? 2 = 8 = 0. 0938 m s? 1 (1M) (1A) 12 (a) Total distance travelled = 60 + 80 + 80 + 60 = 280 m (d) (i) The coin moves in the following sequence: B A C C A B B b) Magnitude of total displacement = 80 + 80 = 160 m 160 m (west). The total displacement of the athlete is Therefore, it is at B finally. zero. the coin is also zero. (1A) (1M) (1A) (1M) (1A) (1M) (1A) (ii) The displacement of the coin is Therefore the average velocity of (c) Total distance travelled = 280 + 60 + 80 = 420 m 14 (a) Total distance = ? r = 5? ? 60 m C = 15. 7 m Total displacement =5+5 = 10 m 80 m  © The total displacement travelled by her is 10 m. (b) Jane’s statement is incorrect. (1A) Since both girls start at X and meet at Y, they have the same displacement. (1A) Betty’s statement is incorrect. 1A) Since both girls return to their starting point, their displacements are zero. (1A) Physics in articles (p. 40) (a) From 19 January 2006 to 28 February 2007, (1A) It takes New Horizons spacecraft a total of 406 days to travel from the Earth to Jupiter. (1A) (b) (i) Average speed total distance travelled = total time of travel (1M) = 8 ? 108 406 ? 24 (1A) (1M) = 8. 21 ? 104 km h? 1 (ii) Average acceleration change in velocity = total time of travel = (8. 23 ? 5. 79)? 10 4 406 ? 24 = 2. 50 ? 104 km h? 2 (1A) (1A) (c) July 2015  © 2 1 2 3 4 5 Motion II 10 (a) The object moves with a constant elocity. Practice 2. 1 (p. 61) D B D D B 30 ? 10 = 10 m s–1 v= 2 (b) The object moves with a uniform acceleration from rest. (c) The object moves with a uniform deceleration, starting with a certain initial velocity. Its velocity becomes zero finally. The velocity of the car at t = 2 s is 10 m s–1. 6 7 C (d) The object first moves with a uniform acceleration from rest, then at a constant velocity, and finally moves with a smaller uniform acceleration again. (a) Total displacement = 4 ? 5 + (? 5) ? (7 ? 5) = 10 m The total displacement from the staircase to her classroom is 10 m. (e) The object moves at a constant velocity and then suddenly moves at constant velocity of same magnitude in the opposite direction. (b) Classroom C 8 (f) The object moves with uniform deceleration from an initial velocity to rest, and continue to move with the uniform acceleration of the same magnitude in opposite direction. 9 (a) The object accelerates. (b) The object first moves with a constant velocity. Then it becomes stationary and finally moves with a higher constant velocity again. 11 (a) The object moves with zero acceleration (with constant velocity of 50 m s–1). (b) The object moves with a uniform cceleration of 5 m s–2. (c) 12 The object moves with uniform deceleration of 5 m s–2. (c) The object decelerates to rest, and then accelerates in opposite direction to return to its starting point. (a) It moves away from the sensor. (d) The object moves with uniform velocity towards the origin (the zero displacement position), passes the origin, and continues to move away from the origin with the same uniform velocity.  © (b) (c) The greatest rate of change in speed 0 ? 3. 5 = 2 = –1. 75 m s–2 (d) Total distance travelled = area under the graph 3. 5 ? 2 2 ? 6 = + 2 2 = 9. 5 m Practice 2. 2 (p. 71) 1 C By v2 = u2 + 2as, 290 3. 6 2 13 (a) =0+2? 1? s s = 3240 m = 3. 24 km < 3. 5 km The minimum length of the runway is 3. 5 km. 2 B Cyclist X is moving at constant speed. Time for cyclist X to reach finish line displacement 150 = = = 30 s time 5 For cyclist Y: u = 5 m s–1, s = 250 m, (b) Total distance travelled = area under the graph (12 + 6) ? 3 = 2 = 27 m a = 2 m s–2 By s = ut + 1 2 at , 2 1 250 = 5 ? t + ? 2 ? t2 2 (c) Average speed total distance travelled = time taken 27 = 3 t = 13. 5 s or t = ? 18. 5 s (rejected) Y needs 13. 5 s to reach finish line. Therefore, cyclist Y will win the race. 3 B Since the bullet start decelerates after fired into the wall, we could just consider the displacement of the bullet in the wall. To prevent the bullet from penetrating the wall, the bullet must stop in the wall. = 9 m s–1 14 (a) She moves towards the motion sensor. (b) The highest speed of the girl in the journey is 3. 5 m s–1.  © By v2 = u2 + 2as, 0 = 500 + 2 ? (? 800 000) ? s 2 8 By v = u + at, 14 = u + 2 ? 5 u = 4 m s–1 s = 0. 156 m = 15. 6 cm < 15. 8 cm The minimum thickness of the wall is 15. 8 m. By v2 = u2 + 2as, 142 = 42 + 2 ? 2 ? s s = 45 m 4 C When the dog catches the thief at t = 5 s, its total displacement is 30 m. The dog is sitting initially, so u = 0. 1 By s = ut + at2, 2 1 30 = 0 + a(5)2 2 The displacement of the girl is 45 m. 9 (a) v = u + at = 0 + 20 ? 0. 3 = 6 m s? 1 The horizontal speed of the ball travelling towards the goalkeeper is 6 m s? 1. a = 2. 4 m s–2 Its acceleration is 2. 4 m s–2. (b) By v2 = u2 + 2as, 02 ? 62 a= = –22. 5 m s? 2 2 ? 0. 8 The deceleration of the football should be 22. 5 m s? 2. 5 6 D 90 36 ? v? u = 3. 6 3. 6 = 1. 5 m s–2 a= t 10 By v = u + 2as, 2 2 10 (a) The reaction time of the cyclist is 0. 5 s. s= v ? u = 2a 2 2 90 3. 6 36 3. 6 2 ? 1. 5 ? 2 2 = 175 m (b) Braking distance (2. ? 0. 5)? 15 = 11. 25 m = 2 Thinking distance = 15 ? 0. 5 = 7. 5 m Stopping distance = 11. 25 + 7. 5 = 18. 75 m child. 20 m The distance travelled by the motorcycle is 175 m and its acceleration is 1. 5 m s . –2 7 (a) Thinking distance = speed ? reaction time 108 = ? 0. 8 = 24 m 3. 6 Therefore, the bicycle would not hit the (b) Since the car decelerates uniformly, braking distance v+u = ? t 2 108 +0 = 3. 6 ? (3 ? 0. 8) 2 = 33 m 11 By v = u2 + 2as, 0 = 32 + 2 ? (–0. 5) ? s s=9m 8m Therefore, the golf ball can reach the hole. 2 12 (a) (i) By v = u + at, 0 = u + (–4)(4. 75) u = 19 m s–1 The initial velocity of the car is 19 m s–1. (c) Stopping distance = thinking distance + braking distance = 24 + 33 = 57 m  © (ii) By v2 = u2 + 2as, 0 = 19 + 2 ? (–4) ? s s = 45. 1 m 2 3 C For option A, apply equation v2 = u2 – 2gs and take s = 0 (the ball returns to the second floor), v = –u = –10 m s–1 (vertically downwards) The displacement of the car before it stops in front of the traffic light is 45. 1 m. This is the same velocity as the initial velocity of option B. Therefore, in both ways the ball has the same vertical speed when it reaches the ground. (b) By v = u + 2as, 17 = 0 + 2 ? 3 ? s s = 48. 2 m 2 2 2 The displacement of the car between starting from rest and moving at 17 m s is 48. 2 m. –1 4 B Take the upward direction as positive. 1 By s = ut + at2, 2 1 0 = u ? 30 + ? (? 10) ? 302 2 u = 150 m s–1 13 (a) By v2 = u2 + 2as, v2 = 0 + 2 ? 0. 1 ? 500 v = 10 m s–1 His speed is 10 m s . –1 (b) Consider the first section. By v = u + at, v? u t= a 10 ? 0 = 0. 1 = 100 s Consider the second section. 1 By s = ut + at2, 2 1 800 = 10t + ? 0. 5t2 2 t = 40 s or t = –80 s (rejected) The speed of the bullet is 150 m s–1 when it is fired. 5 Speed of stone Equation used t=1s t=2s t=3s t=4s v = u + at Distance travelled by the stone 1 s = ut + at 2 2 m 20 m 45 m 80 m 10 m s–1 20 m s 30 m s –1 –1 40 m s–1 Total time taken = 100 + 40 = 140 s It takes 140 s for Jason to travel downhill. 6 1 By s = ut + at2, 2 1 10 = 0 + (10) t2 2 t = 1. 41 s v = u + at Practice 2. 3 (p. 83) 1 2 D D = 0 + 10(1. 41) = 14. 1 m s–1 It takes 1. 41 s for a diver to drop from a 10-m platform. His speed is 14. 1 m s–1 when he enters the water.  © 7 Take the upward direction as positive. By v = u + 2as, 4 = 0 + (2)(–10)s s = 0. 8 m 2 2 2 Besides, since Y spends a shorter time to reach its highest point, it should be fired after X. 10 (a) By s = ut + The highest position reached by the puppy is 0. m above the ground. 8 (a) Consider the boy’s downward journey. Take the downward direction as positive. 1 By s = ut + at2, 2 1 0. 5 = 0 + (10) t2 2 t = 0. 316 s 1 2 at , 2 1 120 = 8t + ? 10 ? t2 2 t = 4. 16 s or t = ? 5. 76 s (rejected) It takes 4. 16 s to reach the ground. (b) v = u + at = 8 + 10 ? 4. 16 = 49. 6 m s–1 Its speed on hitting the ground is 49. 6 m s–1. 11 (a) Distance between the ceiling and her hands = 6 – 2 – 1. 2 = 2. 8 m Hang-time of the boy = 0. 316 ? 2 = 0. 632 s (b) Let s be her vertical displacement when she jumps. As the maximum jumping speed is 8 m s–1, i. e . u = 8 m s–1. By v2 = u2 + 2as, v2 ? 2 s= 2a 2 0 ? 82 = (upwards is positive) 2 ? (? 10) s = 3. 2 m > 2. 8 m Therefore, the indoor playground is not safe for playing trampoline. 1 (a) By s = ut + at2, 2 1 132 = 0 ? t + ? 10 ? t2 2 t = 5. 14 s The vehicle can experience a free fall in the Zero-G facility for 5. 14 s. (b) Take the upward direction as positive. By v = u + 2as, 0 = u + 2 ? (–10) ? 0. 5 u = 3. 16 m s–1 2 2 2 The jumping speed of the boy is 3. 16 m s–1. 9 Take the upward direction as positive. (a) By v2 = u2 + 2as, 0 = u2 + 2(–10)(200) u = 63. 2 m s–1 The velocity of the firework X is 63. 2 m s–1 when it is fired. 12 (b) By v = u + at, = 63. 2 + (–10)t t = 6. 32 s It takes 6. 32 s for the firework X to reach that height. (c) From (a) and (b), for firework Y to explode at 130 m above the ground, the speed of Y should be smaller than that of X. Therefore, Y should be fired at a (b) By v2 = u2 + 2as, v2 = 02 + 2 ? 10 ? 132 v = 51. 4 m s? 1 The speed of the vehicle before it comes to a stop is 51. 4 m s? 1.  © lower speed. (c) Take the upward direction as positive. By v = u + at, –v = v – gt 2v = gt If the stone is projected with a speed of 2v, let the new time of travel be t?. (–2v) = (2v) – gt? v t? = 4 ( ) g = 2t Its new time of travel is 2t. 6 B Take the upward direction as positive. 1 s = ut + at2 2 1 = (10)(4) + (–10)(4)2 2 = –40 m The distance between the sandbag and the ground is 40 m when it leaves the balloon. Revision exercise 2 Multiple-choice (p. 87) 1 D By v2 = u2 + 2as, 0 = 102 + 2a(25 – 10 ? 0. 2) a = –2. 17 m s–2 His minimum deceleration is 2. 17 m s–2. 2 3 D B Consider the rock released from the 2nd floor. By v2 = u2 + 2as, v2 = 2as floor. Note that s2 = 3. 5s. (v2)2 = 2as2 = 3. 5(2as) = 3. 5v2 v2 = 1. 87v (as u = 0) Then consider the rock released from the 7th 7 8 D C Take the downward direction as positive. u = 200 m s–1, v = 5 m s–1, a = ? 0 m s–2 By v = u + at, 5 = 200 + (? 20)t t = 9. 75 s The rockets should be fired for at least 9. 75 s. Both C and D satisfy this requirement. But for D, after firing for 10. 2 s, v = u + at = 200 + (–20)(10. 2) = –4 m s–1 i. e. it flies away from the Moon with 4 m s–1 upwards. It c annot land on the Moon. Therefore, the correct answer is C. 4 5 A C The stone returns to the ground with the same speed (but in opposite direction). 9 10 D D  © 11 12 13 (HKCEE 2006 Paper II Q1) (HKCEE 2007 Paper II Q2) (HKCEE 2007 Paper II Q33) (b) (i) Conventional (p. 89) 1 (a) The reaction time of the driver is 0. 6 s. (b) v a= t = 0 ? 12 3. 6 ? . 6 (1A) (Correct axes with label) from t = 1. 20 s to 1. 25 s) from t = 1. 45 s to 1. 50 s) (1A) (1A) (1A) (A straight line with slope = 0. 35 m s–1 (A straight line with slope = –0. 35 m s–1 (1A) (1M) = –4 m s–2 The acceleration of the car is –4 m s–2. (c) The stopping distance of the car is the area under graph. Stopping distance 12 ? (3. 6 ? 0. 6) =12 ? 0. 6 + 2 = 25. 2 m The stopping distance of the car is shorter than 27 m. The driver will not be charged with driving past a red light. (1A) (1A) (1M) (ii) 2 (a) The object moves away from the motion sensor with uniform velocity at 0. 35 m s–1 from t = 1. 20 s to 1. 25 s. 1A) From t = 1. 25 s to 1. 45 s, the object moves with negative acceleration. (1A) Then, from t = 1. 45 s to 1. 50 s, the object changes its moving direction and moves towards the motion sensor again with a uniform velocity of –0. 35 m s–1. (1A) (Correct axes with labels) (1A) (Correct graph with the acceleration of ? 0. 35 ? 0. 35 about 1. 40 ? 1. 30 = –7 m s–2 at t = 1. 30 s to 1. 40 s) (1A) !  © 3 (a) (b) Total displacement of the car = area bound by the v? t graph and the time axis 1 1 = (5 ? 5) ? (20 ? 3) 2 2 = ? 17. 5 m (1M) (1A) (c) Yes, the car moves 12. 5 m forwards from t = 0 to t = 5 s. Therefore, it hits the roadblock. 1A) 5 Take the upward direction as positive. (a) From point A to the highest point: (Correct axes with labels) (Correct shape of minibus’ graph) (Correct shape of sports car’s graph) (Correct values) (1A) (1A) (1A) (1A) By v2 = u2 + 2as, 0 = 42 + 2 (–10) s s = 0 . 8 m By v = u + at, 0 = 4 + (–10)t t = 0. 4 s (1M) From the highest point to the trampoline: 1 s = ut + at2 (1M) 2 1 = 0 + (–10)(1. 2 – 0. 4)2 2 = –3. 2 m (1A) 3. 2 m above the trampoline. (1A) The maximum height reached by him is (1M) (b) From the graph in (a), the two vehicles have the same velocity at t ? 2. 3 s after passing the traffic light. (1A) (1M) (c) The area under graph is the displacement of the cars. Consider their displacements at t = 3 s, For the sports car: 1 s = ? 15 ? 3 = 22. 5 m 2 For the minibus: 1 s = ? (7 + 13) ? 3 = 30 m 2 The minibus will take the lead 3 s after passing the traffic light. (1A) (b) Height of point A above the trampoline (1A) = 3. 2 – 0. 8 = 2. 4 m (1M) (1A) 6 (a) Initial velocity v = 90 km h–1 90 = m s–1 3. 6 = 25 m s–1 Thinking distance =v? t = 25 ? 0. 2 =5m The thinking distance is 5 m. (1A) (1M) 4 (a) The car moves forward with uniform acceleration at ? 1 m s? 2 from t = 0 s to t = 5 s. (1A) (1A) Then the car changes its moving direction. From t = 5 s to t = 8 s, it moves backwards with a uniform acceleration of ? 6. 67 m s . ?2 Its instantaneous velocity is 0 at t = 5 s. (1A) †  © (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 25 2 = 2 ? (80 ? 5) = ? 4. 17 m s–2 4. 17 m s–2. (1M) (c) The slope of the graph is the magnitude of the acceleration of the apple. speed / m s? 1 7. 75 (1A) (1A) Hence, the deceleration of the car is (c) By v2 = u2 + 2as, s= v ? u 2a 0 2 ? 25 2 = 2 ? ( ? 4. 17 ? 2) 2 2 (1M) 0 0. 775 time / s (Correct labelled axes) (2A) (1A) (Straight line with a slope of 10 m s? 2) = 37. 5 m Braking distance = 37. 5 m Stopping distance = 37. 5 + 5 = 42. m (1M) (d) The two graphs have no difference. (1A) (1A) 8 (a) Take the downward direction as positive. By v2 = u2 + 2gs, v = u + 2 gs 2 The driver could not stop before the traffic light. Therefore, his claim is incorrect. (1A) (1M) 7 (a) Take the downward direction as positive. 1 By s = ut + gt2, 2 1 3 = 0 ? t + ? 10 ? t2 2 3? 2 t= = 0. 775 s 10 (1M) = 0 2 + 2 ? 10 ? (40 ? 3) = 27. 2 m s–1 cushion is 27. 2 m s? 1. 1 (b) (i) By s = ut + gt2, 2 1 40 – 3 = 0 + ? 10 ? t2 2 t = 2. 72 s (1A) The speed of the residents landing on the (1M) (1A) The apple travels in air for 0. 775 s. (1A) (b) By v2 = u2 + 2as, v = 2 ? 10 ? 3 (1M) 1A) –1 = 7. 75 m s? 1 The speed of the apple is 7. 75 m s when the apple just reaches the ground. The time of travel in air is 2. 72 s. u+v (ii) By s = t, (1M) 2 2s t= u+v 2? 3 = t 27. 2 + 0 = 0. 221 s (1A) The time of contact is 0. 221 s.  © (c) (b) Slope of the graph from t = 0 to t = 0. 28 s 2. 3 ? 0 = 0. 28 ? 0 = 8. 21 m s–2 The acceleration of the ball due to gravity is 8. 21 m s–2. (1M) (1A) (c) (Correct labeled axes) (Correct shape) (Correct values) (1A) (1A) (1A) (i) 9 (a) t = 2 s: Displacement of the trolley = 0. 7 ? 0. 15 = 0. 55 m t = 3. 4 s: (1A) Displacement of the trolley = 1. 175 ? 0. 15 = 1. 025 m t = 4. 9 s: 1A) Displacement of the trolley = 0. 6 ? 0 . 15 = 0. 45 m (1A) (b) It moves away from the motion sensor with a changing speed from t = 2 s to t = 3. 4 s. (Correct sign) (Correct shape) (1A) (1A) (1A) (1A) (1A) (ii) The method does not work Then it rests momentarily at t = 3. 4 s. After that, it moves towards the motion since ultrasound will be reflected by the transparent plastic plate. (1A) (c) sensor with a changing speed. 1 By s = ut + at2, 2 1 ? 0. 1 = 0. 7 ? 2. 9 + ? a ? (2. 9)2 2 a = ? 0. 507 m s? 2 (1A) (1M) 11 (a) (i) The ball is held 0. 15 m from sensor before being released. The ball hits the ground which is 1. m from the sensor. (1A) (1A) Therefore, the ball drops a height of 0. 95 m. which are 0. 45 m, 0. 65 m and 0. 775 m from the sensor in its first 3 rebounds. (1A) The acceleration of the trolley is ? 0. 507 m s? 2. (ii) The ball rebounds to the positions 10 (a) The motion sensor is protruded outside the table to avoid the reflection of ultrasonic signal from table. (1A)  © At the 1st rebound, the ball rises up (1. 1 ? 0. 45) = 0. 65 m. nd The average acceleration is 66. 6 m s–2. (1A) (1A) (1A) (c) v / m s? 1 6. 32 At the 2 rebound, the ball rises up (1. 1 ? 0. 65) = 0. 45 m. rd At the 3 rebound, the ball rises up (1. 1 ? 0. 75) = 0. 325 m. (b) (i) The ball hits the ground with velocities of 3. 9 m s , 3. 25 m s and 2. 75 m s–1 in its first 3 rebounds. (3A) 3. 9 (1M) 0. 95 ? 0. 55 (1A) –1 –1 t3 t1 t2 t4 t5 t/s (ii) Acceleration = slope of graph = = 9. 75 m s–2 ?6. 32 (3 straight lines) (Correct slopes) (1A) (1A) 12 Take the downward direction as positive. 1 (a) By s = ut + gt2, (1M) 2 1 2 = 0 ? t + ? 10 ? t2 2 2? 2 t= = 0. 632 s (1A) 10 It takes 0. 632 s from t1 to t2. (Correct labels of time and velocity)(1A) 13 (a) Speed v = 70 km h–1 70 = m s–1 3. 6 = 19. 4 m s–1 d Reaction time = v 6 = 19. 4 = 0. 309 s The reaction time of the man was 0. 09 s. (1M) (b) At t2, v = u + at (1A) = 0 + 10 ? 0. 632 = 6. 32 m s –1 –1 (1 M) Shirley’s speed is 6. 32 m s when she lands on the trampoline at t2. At t4, she leaves the trampoline at the same speed. Therefore, from t3 to t4, by v2 = u2 + 2as, a= v2 ? u2 2s (? 6. 32) 2 ? 0 2 = 2 ? 0. 3 (b) By v2 = u2 + 2as, v2 ? u2 a= 2s 2 0 ? 19. 4 2 = 2 ? 48 = –3. 92 m s–2 3. 92 m s–2. (1M) (1M) (1A) The average deceleration of the car was (c) (1A) Speed v = 80 km h–1 80 = m s–1 3. 6 = 22. 2 m s–1 = 66. 6 m s–2  © Thinking distance = vt = 22. 2 ? 0. 309 = 6. 86 m By v = u + 2as, braking distance s v2 ? u2 = 2a 2 0 ? 22. 2 2 = 2 ? ? 3. 92) 2 2 (1A) Take the upward direction as positive. 1 s = ut + at2 (1M) 2 1 = 7 ? 1. 75 + ? (–10) ? 1. 752 2 = –3. 06 m (negative means the water is below the spring board) The spring board is 3. 06 m above the water. Alternative method: (1A) = 62. 9 m Therefore, the stopping distance = 6. 86 + 62. 9 = 69. 8 m (1A) Consider the upward motion and downward motion separatel y. For the upward motion, she takes 0. 7 s to reach the highest point from the spring board. Take the upward direction as positive. 1 By s = ut + at2, (1M) 2 1 s1 = 7 ? 0. 7 + ? (–10) ? 0. 72 2 = 2. 45 m For the downward motion, she takes 1. 5 s from the highest point to enter water. Take the downward direction as positive. By s = ut + 1 2 gt , 2 1 s2 = 0 + ? 10 ? 1. 052 = 5. 51 m 2 (1A) This stopping distance is greater than the initial distance between the car and the boy. (1A) Therefore, the car would have knocked down the boy if the car had travelled at 80 km h? 1 or faster. (d) A drunk has a longer reaction time. (1A) This means that the thinking distance, and thus the stopping distance (sum of thinking distance and braking distance), increases. (1A) (1M) (1A) 14 (a) Take the upward direction as positive. By v = u + at, u = 0 ? (? 10) ? 0. 7 = 7 m s–1 board is 7 m s . 1 Therefore the height of the spring board above the water = s2 – s1 = 5. 51 – 2. 4 5 = 3. 06 m (1A) (1M) (1A) The speed of Belinda leaving the spring (b) Total time taken from the spring board to the water = 0. 7 + 1. 05 = 1. 75 s (c) v = u + at = 0 + (? 10) ? 1. 05 = ? 10. 5 m s–1 is 10. 5 m s–1.  © The speed of the diver entering the water (d) Deceleration of car Y = slope of the graph during 0. 5 s? 8. 5 s = 0 ? 19. 4 = –2. 43 m s–2 8. 5 ? 0. 5 (1A) The deceleration of car Y is 2. 43 m s–2. (c) Thinking distance = area under the graph during 0? 0. 5 s = 19. 4 ? 0. 5 = 9. 7 m (1A) (Correct shape) (Correct times) (Correct velocities) 1A) (1A) (1A) Braking distance = area under the graph during 0. 5 s? 8. 5 s 1 = ? 19. 4 ? (8. 5 – 0. 5) 2 = 77. 6 m distance are 9. 7 m and 77. 6 m respectively. (1A) The thinking distance and the braking (e) (See the figure in (d). ) (Correct slope – parallel to that in (d). ) (1A) (Correct position – above that in (d). ) (1A) 15 (a) Speed 70 km h–1 70 = m s–1 3 . 6 = 19. 4 m s –1 (d) The coloured area is equal to the difference in the stopping distances travelled by cars X and Y. (1A) (e) (1M) Stopping distance of car X = area under the graph during 0? 5 s 1 = ? 19. 4 ? 5 = 48. 5 m 2 Coloured area = 9. 7 + 77. 6 – 48. = 38. 8 m < 50 m Since the difference in stopping distances of the cars is smaller than the initial separation of the cars, the two cars do not collide with each other before they stop. (1A) (1M) (1M) Distance travelled by car Y in 2 s = vt = 19. 4 ? 2 = 38. 8 m < 50 m Since the distance between the cars is greater than the distance that car Y can travel in 2 s, the driver of car Y obeys the rule. corresponding v–t graph. Deceleration of car X = slope of the graph during 0? 5 s (1A) (1M) (b) Deceleration of a car is the slope of their 0 ? 19. 4 = 5? 0 = –3. 88 m s–2 The deceleration of car X is 3. 88 m s–2. (1A) 16 a) From t = 0 s to t = 5 s, the car moves with a uniform acceleration of 17 ? 0 = 3. 4 m s–2. 5 (1A)  © From t = 5 s to t = 20 s, the car moves with a constant velocity of 17 m s–1. (1A) From t = 20 s to t = 28 s, the car moves with a uniform acceleration of 0 ? 17 = ? 2. 125 m s–2. 28 ? 20 at rest. (1A) (b) s = ut + 1 2 at 2 1 = 0 + ? 17. 5 ? (8 ? 60)2 2 = 2 016 000 m (2016 km) (1M) (1A) The Shuttle travels 2 016 000 m (2016 km) in the first 8 minutes. From t = 28 s to t = 30 s, the car remains (1A) 19 (a) (i) The cyclist is using first gear when the acceleration is greatest before braking. shortest time. (1A) (1A) (1M) (1M) (1A) b) (ii) The cyclist uses second gear for the (b) Distance travelled = area under straight line PQ (8 + 6) ? 2 = 2 = 14 m The cyclist travels 14 m in second gear. (c) The acceleration during t = 18 s? 20 s 0? 9 = (1M) 20 ? 18 = ? 4. 5 m s–2 The deceleration is 4. 5 m s . –2 (1A) (Correct shape) (Correct time instants) (Correct accelerations) (1A) (1A) (1A) (1A) (1A) 20 21 (c) Yes. (HKCEE 2 005 Paper I Q1) 1 (a) s = ut + at2 2 1 = 0 + ? 10 ? (500 ? 10? 3)2 2 = 1. 25 m Therefore the minimum height the (1M) The car changes direction at t = 30 s. Its velocity changes from positive to negative, showing a change in its travelling direction. 1A) (1M) (1A) (1A) laptop must fall for it to be ‘saved’ is 1. 25 m. (b) v = u + at = 0 + 10 ? (500 ? 10 ) = 5 m s? 1 the ground is 5 m s–1. ?3 (1M) (1A) 17 18 (HKCEE 2002 Paper I Q8) (a) v = u + at = 0 + 17. 5 ? 8 ? 60 = 8400 m s–1 minutes is 8400 m s–1. The speed of the computer when it hits The speed of the Shuttle after the first 8  © (c) Most falls are likely to be from below this height, effect. (1A) (1A) (1A) so the protection will not have taken Physics in articles (p. 96) (a) 2. 45 m (b) (i) By v2 = u2 + 2as, u = v ? 2as u2 = 0 ? 2(? 10)(2. 45 + 0. 07 ? 1. 09) u = 5. 35 m s? 1 2 2 (1A) (1M) Take the upward direction as positive. 22 (a) Any one from: Rate of change of displacement Displacement per unit time (1A) (b) The velocity of a braking car is decreasing (with time) (1A) so the car has negative acceleration. (1A) Its displacement is (still) increasing with time, so its velocity is (still) positive In this case, the acceleration and velocity are in opposite directions. (1A) (1A) (1A) The vertical speed of Javier Sotomayor is 5. 35 m s? 1 when he leaves the ground. (ii) Take the upward direction as positive. Consider the upward journey. By v = u + at, v ? u 0 ? 5. 35 t= = = 0. 54 s a ? 10 (1M) (c) i) Consider the downward journey. 1 By s = ut + at2, (1M) 2 1 ? (2. 45 + 0. 07 ? 0. 71) = 0 + (? 10) t2 2 t = 0. 60 s The time that he stays in the air = (0. 54 + 0. 60) = 1. 14 s Alternative method: (1A) (Correct graph) (1A) Take the upward direction as positive. 1 By s = ut + at2, (1M) 2 (0. 71 ? 1. 09) = 5. 35t + 1 (? 10)t 2 (1M) 2 t = 1. 14 s or t = ? 0. 07 s (rejected) (ii) Vertical distance travelled = area under the graph from 4. 0 s to 10. 0 s (70 + 130)? 6 = 2 (1M) (1A) The time that he stays in the air is 1. 14 s. = 600 m (1A) The vertical distance travelled by the rocket between t = 4. 0 s and t = 10. s is 600 m.  © 3 1 2 3 4 C C Force and Motion 6 (a) The MTR train is accelerating in the forward direction. The man tends to move at his original speed (smaller speed), so he would move backwards relative to the MTR train. (b) The MTR train is slowing down. The man tends to move at his original speed (greater speed), so he would move forwards relative to the MTR train. (c) The MTR train is moving forwards at constant velocity. The man moves forwards with the same constant velocity, so he would remain at rest relative to the MTR train. (d) The MTR train is turning a corner. The Practice 3. 1 (p. 104) (b), (e), (f) 5 a) Stretching a rubber band (b) Standing on the floor (c) Walking time (e) (f) A compass A rubbed plastic ruler attracts small bi ts of paper (d) Exists in every object on the earth at any 7 man tends to move at his original direction, so he would move outwards relative to the MTR train. In space, the gravitational force acts on the spaceship is negligible. When the rockets are shut down, they do not exert a force on the spaceship. Therefore, no net force acts on the spaceship. By Newton’s first law, the spaceship is in uniform motion and can travel far out in space. 8 Joan moves on the ice surface with a constant velocity. Practice 3. 2 (p. 111) 1 2 3 4 5 C C D C (a) No. Athletes would hit the wall of the stadium if it is too close to the finishing line. (b) The mat is used to protect the athletes if they hit the wall after passing the finishing line. Practice 3. 3 (p. 122) 1 2 3 4 5 D A B A D  © 6 (a) 7 (a) Horizontal component = 40 + 30 cos 30 ° = 66. 0 N Vertical component = 30 sin 30 ° = 15 N Resultant = 66 2 + 15 2 = 67. 7 N Let ? be the angle between the resultant Resultant’s magnitude is 67 N and the angle between the resultant and the horizontal is 13 °. (b) and the horizontal. 15 tan = ? = 12. 8 ° 66 Resultant’s magnitude is 67. N and the angle between the resultant and the horizontal is 12. 8 °. (b) Horizontal component = 40 + 30 cos 45 ° = 61. 2 N Vertical component = 30 sin 45 ° = 21. 2 N Resultant’s magnitude is 65 N and the angle between the resultant and the horizontal is 19 °. (c) Resultant = 61. 2 2 + 21. 2 2 = 64. 8 N Let ? be the angle between t he resultant and the horizontal. 21. 2 tan = ? = 19. 1 ° 61. 2 Resultant’s magnitude is 64. 8 N and the angle between the resultant and the horizontal is 19. 1 °. (c) Resultant’s magnitude is 60 N and the angle between the resultant and the horizontal is 25 °. (d) Horizontal component = 40 + 30 cos 60 ° = 55 N Vertical component = 30 sin 60 ° = 26. 0 N Resultant = 55 2 + 26. 0 2 = 60. 8 N Let ? be the angle between the resultant and the horizontal. 26. 0 ? = 25. 3 ° tan = 55 Resultant’s magnitude is 60. 8 N and the angle between the resultant and the Resultant’s magnitude is 50 N and the angle between the resultant and the horizontal is 37 °. horizontal is 25. 3 °.  © (d) Resultant = 40 2 + 30 2 = 50 N Let ? be the angle between the resultant and the horizontal. 30 tan = ? = 36. 9 ° 40 Resultant’s magnitude is 50 N and the angle between the resultant and the horizontal is 36. 9 °. Hence, the angle between the two 5-N forces is 120 °. Alternative method: By tip-to-tail method, the two 5-N forces and the resultant 5-N force form an equilateral triangle. It is known that each angle of an equilateral triangle is 60 °. Therefore, the angle between the two 5-N forces is 120 °. 8 (a) 10 (b) Resultant force = 2 ? 400 = 800 N The resultant force provided by the cable is 800 N. 11 For the 2-kg mass: (c) 9 R = weight ? cos ? = 20 cos 30 ° = 17. 3 N Suppose the two forces act in the direction as shown. T = 20 N Therefore we have: Vertical component Fx = 5 sin ? Horizontal component Fy = 5 ? 5 cos ? = 5 ? 1 ? cos ? ) (magnitude of the resultant)2 = Fx2 + Fy 2 52 = (5 sin ? )2 + [5 ? (1 ? cos ? )]2 1 = sin ? + 1 ? 2 cos ? + cos ? 2 2 2T cos 45 ° = W 2 ? 20 ? cos 45 ° = W cos ? = 0. 5 W = 28. 3 N ? = 60 °  © 12 (a) 2T sin 10 ° = 500 T = 1440 N The tension of the string is 1440 N. 3 4 5 6 B C A Net force = ma = 40 ? 0. 5 = 20 N C By v2 – u2 = 2as, 0 à ¢â‚¬â€œ u2 = 2a(20) ? u2 = 40a u2 a=? 40 Resistance = ma = 12 ? ? u2 = –0. 03u2 40 (b) Component of force = T cos 10 ° = 1440 ? cos 10 ° = 1420 N The component of the force that pulls the car is 1420 N. 13 (a) 7 8 ‘A bag of sugar weighs 10 N. ’ or ‘A bag of sugar has a mass of 1 kg. By F = ma, F 800 000 a= = = 2 m s–2 m 4 ? 10 5 (b) As the mass is stationary, the net force acting on it is zero. When it flies horizontally, its acceleration is 2 m s–2. 100 ( )? 0 v? u (a) a = = 3. 6 = 4. 63 m s–2 t 6 The acceleration of the car is 4. 63 m s–2. (c) (i) y-component of F1 = weight of mass = 10 N 9 y-component of F1 = F1 sin 30 ° F1 sin 30 ° = 10 N F1 = 20 N x-component of F1 = F1 cos 30 ° = 20 cos 30 ° = 17. 3 N (b) F = ma = 1500 ? 4. 63 = 6945 N The force provided by the car engine is 6945 N. 10 (a) (ii) y-component of F2 = 0 x-component of F2 = x-component of F1 = 17. 3 N

Social Media Networks in Recruitment

The current changes and advances in technology and the role of internet and social media networks, has provided new opportunities for all organizations to communicate easier and find talented employees with less time and cost. This study is to investigate the impact of using social media networks in recruitment of professional staff at RACK private sector. It will show if the impact of these networks is positive or negative and to what extent does the use of them can lead to effective recruitment to improve the productivity and find highly qualified candidates easily.A axed method is used to collect results about the impact of social media on recruitment. Questionnaires will be administered to human resources directors and managers at RACK private sector companies and random sampling will be utilized. The questionnaires will be distributed randomly among 14 companies and the population will be 40-50 HRS directors/managers. In addition, unstructured Interviews will be used as a second method to emphasize the validity of the results and to increase the sample.Chapter 1 : Study Background Introduction The Emirate of Rasa AAA Kalmia (RACK) Is fast growing and the most northern. RACK has come a destination for many tourist visitors and Investors In the private sector. Many of the light and medium manufacturing as well as service Industries seek on the traditional classified and websites to recruit their employees. At the same time the use of these medium is decreasing as more and more college graduates and researchers are attracted to social media. Private sector employers are now seeing low pool of qualified applicants to fill their Job openings.This resulted in lower performing employees and incompatibility in the workplace. The current changes in technology and the growth in the use of internet and social Edie sites is remarkable. Companies and recruiters, therefore, need to be where the potential candidates are to be able to find the right qualified individuals. This involves engaging with talented people across a wide range of social media networks. Overall, and according to Ponderous, and Olivia, â€Å"social media has improved the recruitment process by making it more democratic and open† (Ponderous & Olives, 2013).Using social media alone however, can negatively impact the relationship building between the companies or HRS professionals and the potential candidates (Raja, 2010). It is therefore difficult to completely replace the traditional recruitment methods by the use of social media in the near future. This paper examines the potential impact which social media may have on the recruitment process. It places emphasis on how private sector corporations and recruitment professionals can benefit from the social media networks to promote their products and services efficiently and target the highly talented and qualified employees.It looks at the role of sites such as Faceable, Linked, and others are playing in the process of re cruiting and hiring professionals. Problem Statement RACK private sector is limited in their recruitment of highly qualified employment by not fully using Social Media as a recruitment tool. Social media can positively influence the recruitment in RACK private sector and can lead to the employment of highly talented and qualified employees. The increase use of social media in other Emirates' may decrease growth in RACK and hinder technological advancement.Most job seekers and employers are using social media so the private sector of RACK needs to address this issue in order to access these Job seekers. Rationale for the Research Many companies seek to find highly qualified employees. As stated by None, companies spend large amount of resources in their recruitment efforts (None, R, 2012). Many of them use various strategies to recruit. Social Media have been successful in many aspects of the career path from networking to marketing specialized goods and services (Ponderous & Olives, 2013).According to Headwords, Social Media is found to be essential among the working population and became the main medium of communication. Many companies, both public and private relied have found success in using social media and have integrate it into their daily operation (Headwords, A. , 2011). This study will show that there will be a significant increase in the recruitment of highly qualified employee in RACK private sector when social media is utilized in the recruitment process. The following research questions will undoubtedly be answered within this research paper: 1 .Which social media sites are mostly used in recruitment in RACK private sector corporations? 2. To what extent does the use of social media networks lead to effective recruitment? 3. What are the advantages and disadvantages of using social media in recruiting individuals? Definition of Terms Social Media is any form of online network that encourage social exchange of ideas, views, personal communication and friendship. Some media develop lifelong relationships and individual collaboration that are informal and transparent.These sites include Faceable, Linked, Video, Sing and Twitter (Ponderous, & Olivia, 2013). Private Sector consists of companies or industries that are independently owned and operated without government interference. Highly Qualified employees are those employees that have high skilled qualifications or possess high educational degrees† (None, 2012). Limitation of the Study The impact of social media on recruitment is a new topic and there is a lack of studies about it. In addition, the study is conducted only in the RACK region.Since confidential information is involved, some information is accepted through verbal interview. The definition of highly qualified is relative to the Job requirements and thus may not be standard across all industries. Although candidates can be reached effectively and easily via social media networks, the risk is, if this strateg y is not accompanied with other traditional recruitment search methods, then some talented candidates may not be targeted because still there are many people who do not use these networks and prefer the traditional way of searching and applying for a Job.Furthermore, online profiles don't certainly give an accurate picture of the individuals. Assessing someone's potential and skills based only on an online profile leaves the door open for unethical practices. Summary The study will show that the use of the Social Media in recruitment in the private sector in RACK improves the applicant pool and enhance the application process. Furthermore the study will show that the result of the use of social media improve productivity and meet the demands for highly qualified employees.The study will also show that social media have a positive impact on recruitment. With the era of social media, the approach to work and find Jobs has changed. Social media will not disappear completely; therefore recruiters and employers can take more benefit of this by adopting hiring and recruitment methods that utilize social media networks. Social media, however, has some limitations associated with its use in recruitment process.Although employers can get some benefits; using social individuals and professionals as well as the impact on the selection criteria of individuals. Instead of identifying social media as a recruitment solution, employers should realize and understand that they need to work more closely with the professionals to be able to get the right candidates they are looking for, rather than spending time, money and other resources without any return or benefit. I personally believe that social media cannot solely replace the traditional recruitment methods in the near future.Chapter 2: Literature review According to Molar (201 1), recruitment can be defined as the process of finding, selecting, attracting and hiring qualified personnel to be employed within an organizatio n and contribute to the achievement of its goals and objectives (Molar, 2011, p. 56). Philips (1999) believes that the recruitment process may involve trying to attract highly qualified and talented individuals, screening the applications, and selecting the right applicant for the Job (Philips, 1999, p. ). He argues that the Recruitment and selection of personnel is considered a very critical component of unman resources functions which drives the organizations' success and development (Philips, 1999, p. 10). Most organizations and companies worldwide use the traditional way of recruiting and some tend to use a mixture of both traditional ways and online recruiting with the use of social media networks (Molar, 2011, p. 68).According to Ponderous and Olives (2013), â€Å"Recruiting e-recruitment, or recruiting via the use of social media networks and internet, is a phenomenon that has led to the appearance of a new market in which there is an unprecedented level of interaction betwe en employers ND potential employees† (Ponderous& Olivia, 2013, p. 33). None (2012), believes that â€Å"online technology and the use of social media in recruiting are crucial to companies that compete for the best talented candidates in a high speed Job market† (None, 2012, p. 77). This is because the use of the social media can save a lot of time, cost and efforts and also allow organizations to target more qualified candidates all over the world (None, 2012, p. 161). This paper aims to discuss the impact of using social media networks in the recruitment and selection process at RACK private sector. Examples of the online social media networks which are used heavily are the use Linked, Faceable, Twitter, What's, etc†¦According to Media (201 1), these social network sites can be defined as â€Å"web-based services that allow individuals to construct a public or semi-public profile within a bounded system, articulate a list of other users with whom they share a co nnection, and view their list of connections† (Media, 2011, p. 13). These network sites nowadays has connected the people all over the world and most individuals are using these sites to search for Jobs besides the other purposes of entertainment and injection with other people (Media, 2011, p. 93).Some organizations had already planned and used these networks in their recruitment process in order to target more qualified individuals worldwide and minimize the cost. These organizations use the social media networks to advertise for their Job openings and at the same Using the social media networks in recruiting employees at RACK private companies may reduce the huge cost that is used in advertising for the Job postings and it can also help RACK private companies to brand their business processes through the use of these sites to market their products and services.RACK private companies nowadays face the challenge of finding the top talented candidates that they need to hire in order to meet the requirement of the business. As suggested by Philips (1999), this problem of finding the right candidates can be solved by finding other strategies of recruiting such as the use of social media networks (Philips, 1999, p. 24). According to Molar (201 1), there is an increase trend internationally to use the corporate websites and social networks in recruiting and selecting individuals.This can open the chances for RACK private companies to target more international antedates from all over the world with the availability of these sites that connect all people at a minimum cost (Molar, 2011, p. 256). Furthermore, by using these social media sites, the vacancies can be filled faster and therefore save the time that can be spent in searching for qualified candidates through the uses of traditional methods of advertising and Job posting (Headwords, 2011, p. 11).Headwords (2011) argues that these sites can also help organizations to increase their brand visibility onlin e which can establish an excellent image and brand for these organizations. (Headwords, 2011, p. 118). Therefore, RACK private companies can use the social media networks to brand their products and services and brand the companies' image worldwide. By using these sites private companies at RACK can post their vacancies to a larger community and their postings can be accessed by a larger number of qualified candidates (Headwords, 2011, p. 45). This can help private companies at RACK to increase the quality of their hires by attracting the right people for the right Jobs through these social media networks (Sweeney, 2011, p. 58). In addition, there are other benefits of using social media and according to Ponderous and Olivia (2013), â€Å"social media has improved the recruitment process by making it more democratic and open† (Ponderous & Olives, 2013, p. 74).So, private companies at RACK can benefit from the use of these social networks to make its vacancies and Job posting open internationally to all people and this will help immensely to have a wide pool of applicants where it will be easier to find talented potentials. Method This study intends to investigate the impact of using social media on recruitment process and the perception of the HRS professionals at RACK private sector about the SE of these social networks in hiring and selecting personnel.A mixed method will be utilized for this research which consists of both quantitative and qualitative data collection tools in order to provide more in depth data collection and ensure more accurate result of the impact of social media on recruitment process. Participants The population of this study consists of all directors and HRS managers of private companies in Rasa AAA Shaman. In order to answer the research questions, a total of 40-50 respondents from 20 companies in Rasa AAA Shaman private sector were elected based on a random or probability sampling, so all participants will have equal opportun ity to take part in this research.Selected participants will answer a questionnaire structured in Liker format to ask about the use and the impact of social media on recruitment process. The data that will be collected from the respondents will be calculated for clarification and analysis. Instruments A survey questionnaire using the Liker format will be used in this research . The scale below will be used to analyses the answers of all the respondents for each question by calculating the weighted mean: Range Interpretation 3. 01 – 4. 00 Agree 1. 01 -2. 00 Strongly Agree 2. 1 – 3. 00 Disagree 0. 00- 1. 00 Strongly Disagree To test the validity of the questionnaire, it will distributed to 5 participants and these participants along with their results of the questionnaire, will not be part of the research and they were used Just to test the validity. Questionnaire development Seven questions will be used to determine the possible relationship between the variables (socia l media and recruitment) . The questions that were included in the questionnaire had a choice of 4 points from strongly agree to strongly disagree.All questions were selected carefully to be relevant to the topic and to reflect the aim of this study and enable to get the right results. The questionnaire will have the following topics: 1. The usage of internet and social networks to target talented candidates. 2. Participants' opinion about the idea of using social media in recruitment (open? Disagree? ) 3. The support from the top management for the use of social media in recruitment. 4. The benefits of using social media in recruitment such as saving time, cost, etc.. 5.The role of social media in branding and marketing of organizations 6. Disadvantages of using social media in recruitment. 7. The importance of using the traditional ways of recruiting besides the use of social media. Data Collection Plan Primary research and secondary research will be used. The primary research wil l be meetings, observation and general discussions with those directors will be used. The secondary data is based on the literature review including articles, Journals and books which was collected earlier about the impact of using social media in recruitment process.Statistical analysis of the data When the questionnaire will be collected from the participants, statistics will be used o analyze all the data through the use of SPAS to come up with the statistical analysis for this study. The aim of this study was to identify the impact of using social media (independent variable) on recruitment process (dependent variable). For the purpose of testing the hypothesis, analysis of data will be done and represented in tables.HI will indicate that there is a positive relationship and good impact of using social media in recruiting. H2O will show that there is a negative impact of social media networks if used in recruitment process. HO will indicate that there is no impact or any relatio nship between the use of social media and recruitment process. Implication and Limitation Studies about the use of social media networks are away from its impact on recruitment and the effective use of it in hiring as HRS function.The conclusion which this research will be able to draw, is the how HRS professionals can benefit from the use of social media and what are pros and cons and the impact of social networks if applied and used in recruitment instead of the other traditional methods of hiring. This research will also reveal that social media networks can be used in other littorals and not only recruitment; it can be used for example in marketing campaigns and branding. This research will be conducted in RACK region only and the results may not apply to other companies in the AAU or may not benefit them.Some of the questions that will be asked to participants, considered to be confidential and therefore respondents may not provide the accurate answer and this will affect the f indings and results of the research. Appendices and References Appendix A: Questionnaire 1. The use of internet and social networks can improve the way we target talented candidates from all over the world. Strongly Agree 4 Agree 3 Disagree 2 Strongly disagree 2. I am very open to the idea of using these social media networks fully in the recruitment process without the need to use the traditional methods of hiring. Ring employees 4. Social networks when used in recruiting can save time and reduce the cost of advertisement for Job postings 5. Social media can contribute to the branding of the organization if used in the right way cannot trust the social media users because some conditional information cannot be published online and this therefore can affect the selection decision.

Music: Native Americans in the United States and Correct Answer Essay

Incorrect 1. The language used to discuss music is universal. A) True B) False Table for Individual Question Feedback Points Earned: 0. 0/4. 0 Correct Answer(s): B Correct 2. One way 19th-century Americans were exposed to European classical music was through visiting European musicians. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 3. The musical language of America is based on: A) Western European musical concepts B) ancient Greek musical concepts C) neither a nor b D) a and b Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 4. When categorizing music, stylistic labels should be adhered to rigidly. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): B Correct 5. Diverse musical traditions are a contributing element to our rich national culture. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 6. The composer of a folk song is often: A) poor B) not a musician C) old D) unknown Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): D. Correct 7. Early American settlers _______________ the music of Native Americans. A) accepted B) tolerated C) copied D) rejected Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): D Correct 8. Most early religious, folk, and popular songs were derived the styles of: A) the British Isles B) South America C) a blending of Native American and European cultures D) Native American peoples Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 9. The instruments Americans have are, for the most part, derived from those in. Europe and the British Isles. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 10. Ethnomusicologists study the social aspects of music. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 11. Generally, a culture’s belief systems and music are separate. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): B Correct 12. The merging of cultural traditions produces: A) very little of national value B) ethnic minorities. C) new styles and modes of behavior D) race tensions Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): C Incorrect 13. It is necessary for professional musicians to earn the bulk of their living from performing. A) True B) False Table for Individual Question Feedback Points Earned: 0. 0/4. 0 Correct Answer(s): B Correct 14. The primary factor in the development of America’s cultural mainstream was the predominance of: A) English-speaking settlers B) African slaves C) Native American culture D) free trade Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 15. The __________ Law of 1909 provided ownership of popular songs to composers, lyricists, and publishers. A) Permission B) Copyright C) Ownership D) Holder Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): B Correct 16. Ethnomusicologists study audiences as well as performers. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 17. Many cultures in the world are still unaffected by outside influences. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): B Correct 18. Scientific research has found that music powerfully affects the emotions of listeners. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A Correct 19. Ethnomusicologists rely on a ______________ rather than an ethnocentric perspective to study and describe music. A) regional B) local C) global D) universal Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): C Correct 20. Who composed music for orchestra, but in the â€Å"pop† style? A) Bob Dylan B) Tchaikovsky. C) George Gershwin D) Beethoven Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): C Correct 21. According to the text, it is unnecessary to acquire a sense of what exists beyond our own cultural experience. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): B Correct 22. When music is transmitted from generation to generation by imitation or memory, this is referred to as: A) notational tradition B) oral tradition C) familial tradition D) none of these Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): B Correct 23. Music labels can be: A) helpful B) ambiguous C) a and b D) neither a nor b Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): C Correct 24. True artistry is generally only found in classical music. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): B Correct 25. Music has taken on a global perspective due to technological advances in communication and transportation. A) True B) False Table for Individual Question Feedback Points Earned: 4. 0/4. 0 Correct Answer(s): A.

Lvmh in the Recession the Substance of Style

http://www. economist. com/node/14447276 LVMH in the recession The substance of style The world’s biggest luxury-goods group is benefiting from a flight to quality, but the recession is also prompting questions about the company’s breadth and balance Sep 17th 2009 | Paris | from the print edition * * Bloomberg â€Å"THERE are four main elements to our business model—product, distribution, communication and price,† explains an executive at LVMH, the world's largest luxury-goods group. â€Å"Our job is to do such a fantastic job on the first three that people forget all about the fourth. For decades LVMH's formula has worked like a spell: seduced by beautiful status-symbols, perfect shops and clever advertising, millions of people have swooned forgetfully towards the firm's cash registers. At Louis Vuitton, LVMH's star company, the model's pricing power has yielded consistent profit margins of around 40-45%, the highest of any luxury-goods brand. These days customers are finding it far harder to forget about price. The seriously rich, of course, are still spending freely.But much of the industry's rapid growth in the past decade came from middle-class people, often buying on credit or on the back of rising house prices. According to Luca Solca of Bernstein Research, 60% of the luxury market is now based on demand from â€Å"aspirational† customers rather than from the wealthy elite. The recession has quickly reversed the trend to trade up, and people are delaying expensive purchases. Bain & Company, a consulting firm, expects the industry's sales to fall by a tenth in 2009, to â‚ ¬153 billion ($225 billion).Some executives even expect a lasting shift in customers' preferences, towards discretion and value. Bernard Arnault, chairman and chief executive of LVMH, believes that the whole industry needs to rebrand itself. â€Å"The word luxury suggests triviality and showing off, and the time for all that has gone,† he say s. Brands which sold â€Å"blingy† easy-to-sell products, milking old names, he says, will fare particularly badly in the new environment. LVMH, by contrast, has never taken such an approach, he says, instead emphasising quality, innovation and creativity.To underline these values, the group is going back to basics in its daily operations. â€Å"Before the crisis, we were putting a lot of energy into beautiful stores, but now we care a bit less about expanding our network and even more about design and price,† says an executive. A few years ago, for instance, at the height of the boom, one LVMH brand was putting diamonds all over its watches, so that it was almost difficult to tell the time. â€Å"Now we are getting back to what really matters, which is nice movements and design,† he says.For some luxury firms, the recession's effects have already been brutal. Private-equity firms and other outside investors which rushed into the industry at its peak have suffer ed most. â€Å"At the top of the market this industry was perceived as easy by outsiders,† says Mr Arnault. â€Å"You borrowed 80% of a target's asking price and hired a good designer, but the strategy has not been successful in several cases. † Lenders to Valentino, an Italian fashion house, are reportedly trying to renegotiate its debt. Permira, a private-equity group, bought the firm in 2007 in a deal valuing it at â‚ ¬5. billion. Permira has since written down its equity investment of about â‚ ¬900m by more than half. Prada Holding, through which Miuccia Prada and her husband control Prada Group, another Italian house, recently restructured its loans in order to defer payment to banks. Prada Group has denied that there are talks to bring in a minority shareholder. Two particularly weak firms, Christian Lacroix, a Paris-based ready-to-wear and haute couture label which used to be part of LVMH, and Escada, a German maker of luxury womenswear, filed for bankrupt cy earlier this year.Amid this turmoil, LVMH is performing relatively well (see chart 1). It has benefited from an established pattern in the luxury industry: when people have less, they spend what they do have on the best quality. Shoppers are going for fewer, classic items—one Burberry raincoat, rather than three designer dresses, or a single Kelly bag by Hermes, a French luxury-goods group, instead of four bags from various lesser designers. For this reason, says Yves Carcelle, chief executive of Louis Vuitton and president of fashion and leather goods for LVMH, â€Å"Vuitton always gains market share in crises. As reliable and sturdy as one of its own handbags, therefore, Vuitton is carrying LVMH fairly comfortably through the recession. In the first half of 2009 the group's revenues were about the same as a year before, though profits were 12% lower. Two divisions—wine and spirits, and watches and jewellery—were the worst affected: their revenues each fell by 17% and their profits by 41% and 73% respectively (see chart 2). Rapid de-stocking by retailers exacerbated the effect of falling demand.But the falls were offset by Vuitton, where revenue rose by a double-digit percentage, registering gains in every market. â€Å"It is incredible that in a downturn the consumer still buys so many Louis Vuitton bags, but she or he does,† says Melanie Flouquet, luxury-goods analyst at JPMorgan in Paris. Vuitton's performance, and the overall robustness of LVMH, a global conglomerate with more than 50 brands and revenues of â‚ ¬17. 2 billion in 2008, should allow it to take advantage of its competitors' weakness in the recession. In the next few years we expect several failures in the industry and good opportunities to acquire assets at attractive prices,† says Mr Arnault. Shareholders in the firm are particularly preoccupied by what he might buy and sell in the next few years. What explains Vuitton's resilience? Beneath the gloss of advertising campaigns, catwalk shows and each season's fleeting trends, Vuitton brings a machine-like discipline to the selling of fancy leather goods and fashion. It is the only leather-goods firm, for instance, which never puts its products on sale at a discount.It destroys stock instead, keeping a close eye on the proportion it ends up scrapping (which it calls the â€Å"destruction margin†). In 2005, when Maurizio Borletti, owner of several prominent department stores in Italy and France, was preparing for the opening of a refurbished La Rinascente department store in Milan, he recalls, the Vuitton people built a scale model of the building in their offices to understand customer flows and get the best positioning. â€Å"In this they're the most professional in the industry,† he says.Unlike most other luxury marques, Vuitton never gives licences to outside firms, to avoid brand degradation. Its factories use techniques from other industries, notably carmaking, t o push costs down ruthlessly and to allow teams of workers to be switched from one product to another as demand dictates. It has adopted methods of quality control, too: one quality supervisor came from Valeo, a French auto-parts supplier. The result is long-lasting utility, beyond show, which is valuable in difficult times. Owning shops gives Vuitton control over levels of stock, presentation and pricing.It was not therefore affected by the panicked price-slashing of up to 80% by American luxury department stores in the run-up to Christmas last year—a â€Å"catastrophe† for others in the industry, according to Mr Arnault. Although other LVMH divisions have been hit by outside retailers de-stocking during the crisis, Vuitton has managed its own inventory, with no competition for space from other brands. With a global network, says Mr Carcelle, the firm can move poorly selling stock to shops where it has performed better. The luxury of diversityVuitton's ability to offs et the steep falls in other divisions shows the value of the diversified conglomerate model in luxury goods. Richemont, the industry's second-largest company, has a less varied portfolio and greater exposure to watches and jewellery, demand for which has been especially weak. According to a recent trading statement, its sales fell by 16% in the five months to the end of August. A group structure also yields savings when negotiating deals for advertising space, property and credit-card fees. It helps to have a specialist beauty retailer, Sephora, and a chain of airport shops, DFS, to sell perfumes and cosmetics.When Vuitton develops watches, say, it can call on the talents of TAG Heuer. But LVMH's breadth also comes in for criticism. Although there is undoubtedly value in some diversification, some people ask whether 50-odd brands under one roof are too many. Vuitton, for instance, would doubtless like to see disposals of weaker brands as a result of the crisis, and a greater concent ration of resources on the group's key businesses. The group's executives devote the bulk of their attention to the most important of these: Louis Vuitton, Moet Hennessy in drinks, TAG Heuer in watches, Christian Dior in perfumes and cosmetics, Sephora and DFS.The group has many smaller businesses, and these get much less attention in such a big group. LVMH does not disclose financial figures for individual brands, but at its presentation of first-half results the group's finance director replied to an analyst asking about fashion and leather-goods that a â€Å"handful† had lost money â€Å"somewhere†. There is speculation that Celine, a ready-to-wear clothing and accessories label, Kenzo, a fashion brand which analysts have long suggested LVMH dispose of, or Loewe, a Spanish leather-goods brand which has so far failed o win much of a following outside Spain and Japan, are among the less profitable. Nevertheless, the group can use the might of Vuitton to support its sm aller, upcoming brands. A department store, for instance, may be asked to take Loewe or Celine in order to get Vuitton. That often frustrates people at Vuitton, however, who would prefer to use the power of the brand for its own benefit, says a person who knows the company well. â€Å"They've never heard of another of LVMH's brands saying, ‘Either give this to Vuitton or I won't come',† he says.Apart from the synergy in watch design, Vuitton does not find that it benefits much from the rest of the group. The reason why LVMH has many small brands which aren't quite making it, says another person familiar with the company, is that Mr Arnault is an optimist who believes that every property can at some point be turned around. That can pay off: some years ago Mr Arnault halted the imminent sale of a make-up line. Thanks to the distribution muscle of Sephora, it has since turned into a bestseller in America.Investors, however, are nevertheless wary of what they see as Mr Arna ult's tendency to collect brands. The crisis has also underlined the fact that Vuitton dominates the group's results. Were it not for Vuitton, estimates one analyst, LVMH's sales would have fallen by 3% in the first half of 2009 and profits would have plunged by 40%. In normal times Vuitton contributes about half of the group's profits, and most of the rest comes from Moet Hennessy. In the first half of this year, however, Vuitton contributed an estimated 70% of profit.That leads some people to question whether LVMH is overly dependent on the leather-goods firm. â€Å"You can argue that there's nothing as good as Vuitton in LVMH's portfolio,† says Pierre Mallevays of Savigny Partners, who was formerly director of acquisitions at LVMH, â€Å"but that simply states the fact that LV's business model is the gold standard of luxury brands; no other brand in the world compares to it. † The biggest risk to LVMH is Vuitton, argues Ms Flouquet, since it accounts for such a big proportion of profits; the company depends on it, she says.The risk to Vuitton, in turn, is that it could fall out of fashion or lose its exclusivity in the eyes of consumers. So far there is no sign of fatigue with the brand. LVMH's senior managers have devised ways to refresh it. In the late 1990s, for example, Mr Arnault saw that there was a risk that as a maker of leather goods alone, Vuitton could be perceived as boring. In 1997 he hired Marc Jacobs, then a relatively unknown designer, to design a fashion line. The aim was to generate seasonal buzz and press coverage.Vuitton's senior executives at the time were against the idea, fearing that adding fashion could undermine a timeless image, but Mr Arnault's move proved successful. To avoid overexposure of its signature â€Å"Monogram† print, Vuitton has taken care to develop a wide range of products and other patterns. â€Å"We increase the number of product lines and we are careful to have several different colours and shapes,† says Mr Arnault. Thus Vuitton sells reasonably priced handbags—the smallest Speedy Bag costs â‚ ¬430 in Paris—but also wildly expensive custom-made luggage, reinforcing its exclusive image.Another effective tactic is to make limited-edition handbags which are hard to get hold of. Five or so years ago Vuitton depended to a large degree on one market, Japan. Most Japanese women owned at least one Vuitton product—and hence provided a large proportion of Vuitton's profits, which worried analysts at the time. Yet the Japanese market for luxury goods was souring. Spending on such items in Japan has fallen sharply since the end of 2005, according to a recent report by McKinsey, a consulting firm. Young women are more individualistic than their mothers, and are seeking out lesser-known brands. You used to see thousands of Vuitton bags coming at you in the Ginza shopping district but far fewer now,† says Radha Chadha, author of a book, â€Å"The Cult of the Luxury Brand: Inside Asia's Love Affair with Luxury†. That reliance on one country is no longer so marked (see chart 3). Fortunately, Vuitton has since rapidly established a strong position in what it hopes will become another Japan: China. â€Å"The Chinese consumer is in a love affair with the Vuitton brand,† says Ms Flouquet. According to LVMH, in the first half of 2009 sales to Chinese people (at home and travelling) made up 18% of Vuitton's revenue.Despite widespread concerns about counterfeiting in the country, the Chinese are now Vuitton's biggest customer base after the Japanese. The key to the firm's success, says Mr Arnault, has been approaching the market exactly as if it were a developed market. â€Å"We treat the Chinese customer as being very sophisticated. † Many competitors, by contrast, have at times lowered their standards for shops in China, he says, using inferior furniture or positioning their stores poorly. Going into new markets and developing new product lines will enable Vuitton o continue producing double-digit growth for years to come, says Mr Carcelle. On every trip to mainland China—he makes five or six a year—he tries to discover a new city and meet its mayor. Mr Carcelle is also tackling other new frontiers: in October he will open a shop in Sukhbaatar Square in Ulan Bator. â€Å"Already if you go to an upmarket disco in Ulan Bator you will see a significant number of our bags,† he says. Vuitton's expansion into China, Mongolia and new product lines such as watches and shoes, suggest that the leather-goods firm will continue to be LVMH's main source of growth.However, it also means that the group may become more rather than less reliant on Vuitton. In theory, the answer could lie in strengthening some of LVMH's smaller names, such as Fendi, a fashion and leather-goods brand. But buying a big, established, global brand with potential for growth could be both a quicker and a sure r route. Or maybe that oneImagineChina A new collection? Analysts and bankers are convinced that Mr Arnault wants to buy the Hermes Group, a producer of leather goods and fashion which matches Vuitton for quality and design.Because Hermes is run so conservatively, says an investment banker who knows LVMH well, it is only a quarter of the size that it could be. â€Å"Mr Arnault would grow it while preserving its values,† he says. Earlier this year, there were rumours that LVMH would sell Moet Hennessy to Diageo, the world's biggest spirits group, which already owns 34% of the business. Such a sale could raise money to buy Hermes. Mr Arnault, however, refuses to be drawn into commenting. For the moment, such an acquisition is impossible, since the family which controls Hermes does not want to sell, and the firm is strongly defended against takeover.Nevertheless, says the banker, the family which controls it has several branches, all with different views. â€Å"It's a pressure cooker and some day it will blow up,† he says. Chanel, another closely held global luxury brand, could also make a desirable target for LVMH. Some people recommend a merger with Richemont, which, Mr Solca argues, would address LVMH's relative weakness in watches and jewellery. Any such deals, or selling Moet Hennessy, would radically change the balance of the group. â€Å"I would be surprised if LVMH sold Moet Hennessy. The business has high margins, high ashflow and it is well managed,† says Ms Flouquet. â€Å"They would probably only sell it if they had a large deal ahead. † Shareholders are nervous that LVMH will pay too high a price for a large acquisition. For this reason the group's valuation may not fully reflect its performance during the crisis. Such concerns are not likely to deter Mr Arnault, who has demonstrated his confidence in LVMH's prospects in luxury by raising his stake in the group over time: he owns 47%. If LVMH does go shopping, it will prob ably behave like one of its best customers: with price in mind, but willing to spend on enduring prestige.

Justification for an Internal Control System

Over the past years many organizations have fallen because of inadequate financial reporting and ineffective controls. To overcome this dilemma, the creation of the Sarbanes Oxley Act (SOX) of 2002 requires corporations to take full control over its financial reporting and accounting by placing internal controls within its organization. Internal controls not only establish the foundation of reasonable assurance for meeting company objectives but also provide functions in achieving other objectives. These objectives are operational effectiveness and efficiency, relevant and reliable financial data, and verify law and regulation compliance. As a controller of this company one believes that internal controls are important for these areas to be successful. Although this business uses the insurance and portfolio approaches as controls to manage the association of risks with activities, one believes an internal control system would be more beneficial for this company. Current Approaches Insurance Approach Insurance is not as large as it can be for a company because it is a way of looking at the risk along with knowing that an acceptance of loss is present for a company. Companies basically carry insurance policies to prevent and cover large liabilities from natural disasters or accidents. Under this approach management is stating an acceptance of a loss is present for a company when insuring the company, assets, or its employees. For a company using this approach protection is the only gain in knowing the company has insurance to make a claim to replace or receive monies for the loss (McCarthy & Flynn, 2004, p. 75). However, the insurance approach is more of a tool for risk financing than a tool for risk management. This is because this approach is reactive in mitigating the impact of a loss rather than preventive in protecting the company from a loss. This approach can be beneficial for the company if a company can find affordable insurance with deductibles the company feels comfortable with. This is because the company would only be responsible for paying the deductible and the insurance company would pay the rest in the event of a catastrophic disaster. Portfolio Approach Unlike, the insurance approach the portfolio approach has more structure and complexity. This is because during the process of decision-making it gives more procedures and processes in making a decision (Thomas, 2002, para. 23). The main idea behind this approach is to maximize investments of a company while minimizing the risks of the company. Even though this approach organizes to an extent the decision-making process it does not provide protection assurance for the investment against risk. The portfolio approach is beneficial for a company when the company wants to measure the type of risk it wants to take on along with the likelihood of making a positive return on that risk. Internal Control System To an organization a vital component to risk management procedures is a reliable internal control system. This system helps regulate, reduce lost, and minimize risks along with accomplishing the organizational goals and success of a company (McCarthy & Flynn, 2004, p. 249). Benefits of an Internal Control System Generally the insurance approach is necessary for a company to overcome the risk of a loss. The portfolio approach is an effective approach but is more reactive than preventive. Even though a business may have insurance and portfolio approaches in place these approaches are not efficient and cost-effective enough to protect the company from risks like an internal control system can. Internal control systems are unlike the insurance and portfolio approaches because these systems are proactive tools in risk management. This type of system ensures the protection of company assets through a system of policies and procedures. In addition, this system establishes reliability in financial data along with establishes compliance with laws and regulations set forth from regulations like the SOX Act. These types of systems also help to improve internal and external communication processes within a company. Recommendation As a controller of this company, one recommends that management incorporates an internal control system into the company. This is because this system will be more beneficial to the company in the long-run than the current approaches the company already has in place. One believes an internal control system will help protect the company from uncertainties as well as ensuring the company is operating in proper accordance with its mission and goals. Justification for an Internal Control System The internal control system has been used since the company was in need of the system and until this day it has been working to its fullest potential. Internal control plays an important key in making sure that the accounting information, financial data, meeting the targets, and ensuring that the management policies are getting followed. There are two elements in making an internal control system successful. These elements are portfolio approaches and insurance. Even though they both help the internal controls, they are somewhat different.Portfolio approach is used in different ways, this helps make investments decisions easier. It also balances the risk against the routine of the company. When discussing portfolio management it is known that there are two types of management: active and passive. Active management can be only one manager or a team but regardless if it one or more. They all have the same idea in mind, which is to get a better market return and they do this by constant ly checking the funds portfolio. A passive management just checks the market index; it does not necessarily say that the passive management is less capable of doing its job.â€Å"Every company's risk management â€Å"solution† will be unique because the exposures and risk appetites all differ. The key is to have a reasonable under-standing of how each treatment option works, alone, and in combination with others, so that decisions are informed and results are less influenced by luck than by reason (McCarthy, Flynn, & Brownstein, 2004). The appetite for risk will always depend on the management team. We will need to understand every risk and think of the options before continuing. A great return is always good but a big loss will hurt more.Insurance is another element that was put in place with the internal controls. Insurance will protect the company in case of an error occurs. There can always be risks in a company, but it is the way we handle them, what is important. When they add insurance it is for a peace of mind, a company wants to be cover in case something did happen. Weather risks can happen anytime and any day. It can be challenging thinking best option for the business. Due to the fact that no one can know what will happen tomorrow but is it better to be protected. As the controller some of the aspects to look into is what do we need.There hasn’t been a tornado in this area for more than 50 years, the question to think is, and do we need tornado insurance? The company was built once but if a risk strikes, it would be really hard to restart the company without developing a financial plan. The company will continue to grow every day, and we need to keep that in mind with the insurance. When the insurance was first put in place, the management team covers everything they thought was needed. However, we may not need certain things that are currently been covered by the insurance. That is why it is very important to do a six month or even a year checkup on the insurance plan.As the controller of the company there are tasks to be completed such as compliance, reporting, budgets, analyzing, and goals. The internal controls help the company and especially the controller achieve all these tasks and stay up-to-date with them. Having internal controls can prevent any losses due to fraud and minimize the loss in assets. It also helps with everyday business activities and what to do in a situation in which a risk is encounter. For the previous reasons that were discuss the company’s success will be much better off having the internal controls with a combination of insurance and portfolio approaches.