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you 1 a couple of 3 C Motion I 7 (a) From you January 2009 to 12 January 2009, the watch runs slower compared to the actual time by on the lookout for minutes. Therefore , when the real time can be 2: 00 pm about 10 January 2009, enough time shown within the watch ought to be 1: 51 pm about 10 January 2009. Practice 1 .

1 (p. 6) D (a) Possible percentage error 10? 6 =? 100% 24? 3600 = 1 . of sixteen? 10 % you (b) sama dengan 1 000 000 times 10? 6 “9 It will take 1 000 500 days being in error by one particular s. (b) Percentage mistake 9 sama dengan? 100% on the lookout for? 24? sixty = six. 94? 10″2% 4 (a) One day sama dengan 24? 70? 60 = 86 400 s Practice 1 . 2 (p. 15) 1 a couple of 3 4 5 C B G D (b) One year = 365? eighty six 400 = 31 five-hundred 000 t 5

Allow t end up being the period of time recorded by a stop-watch. Percentage mistake = zero. 4? completely? 1% t t? 45 s (a) Total length she trips 2? 10 2? 20 2? 12-15 + + = 2 2 2 = 141 m (b) Magnitude of total displacement = 12? 2 + 20? a couple of + 15? 2 = 90 meters Direction: east Her total displacement is 90 meters east. The minimum time frame is forty s. six (a) Percentage error error due to reaction time sama dengan? 100% time measured 0. 3 =? 100% twelve = 3% 6 7 His total displacement can be 0. While using notation inside the figure listed below. (b) By (a), the proportion error of a short time period (e. g. 10 s) measured by a stop-watch is incredibly large.

Because the time periods of 110-m hurdles are incredibly short inside the Olympic Games, stop-watches are not used to avoid significant percentage problems. Since ZX = ZY = you m,? sama dengan? = 60. Therefore , XY = ZX = ZY = 1 m The magnitude in the displacement in the ball can be 1 meters. almost 8 (a) The length travelled by ball will probably be longer whether it takes a rounded path. several (a) Length of the path sama dengan 0. almost eight? 120 = 96 meters (b) No matter which path the ball takes, its shift remains the same. (b) Length of AB over the dotted line ninety six = 40. 6 m = (c) Magnitude of Jack’s average velocity 40. 6? a couple of = = 0. fifty-one m s”1 120 Practice 1 . 3 (p. 23) 1

B Total period 5000 5000 = & = 9821 s 1 . 4 zero. 8 5000 + 5000 = 1 ) 02 meters s”1 Common speed = 9821 Practice 1 . 4 (p. 31) 1 a couple of C W Final rate = 1 ) 5? you ” zero. 2? you = 1 . 3 m s”1 2 C Total time = 9821 & 10? sixty =10 421 s 5000 + 5000 Average acceleration = sama dengan 0. ninety six m s”1 10 421 3 A By a = 3 G When the spacecraft had just finished one particular revolution, the spacecraft came back to it is starting point. Consequently , its displacement was actually zero and its average velocity was also no. v? u, t sixth is v = u + by 36 sama dengan + (? 1 . 5)? 2 3. 6 sama dengan 7 m s”1 sama dengan 7? several. 6 kilometres h”1 = 25. two km h”1 Its velocity after two s is usually 25. a couple of km h”1. 4 your five D (a) Average speed 100 = = 10. m s”1 9. 69 (b) Yes. This is because the magnitude in the displacement is definitely equal to the distance in this case. four B Take those direction from the original way as confident. Average speeding of the ball? 10? 17 = zero. 8 = “33. 8 m s”2 The value of the normal acceleration with the ball can be 33. almost eight m s”2. v? u By a sama dengan, t 100? 0 sixth is v? u three or more. 6 t= = = 4. twenty-seven s a 6. 5 6 (a) Two automobiles move with all the same acceleration, e. g. 50 kilometers h”1, but also in opposite guidelines. (b) A man runs in regards to 400-m recreation space. When we determine his average speed, we could take 400 m since the distance great average acceleration is non-zero.

But as his shift is zero (he comes back to his starting point), his common velocity can be zero. a few The least time it will take is 4. 27 s. six Time as well as s “1 4 0 2 5 6 seventeen 8 twenty two D Normal speed 85 + sixty = five = twenty-eight km h”1 Average speed = Acceleration / m s a couple of 7 doze v? u 22? 2 a= sama dengan 2 . 5 m s”2 = t 8 The acceleration in the car is definitely 2 . 5 m s”2. 7 (a) I will choose ‘towards the left’ while the positive course. 80 two + sixty 2 a few (b) a few = twenty km h”1 C Total time 15 10 = + a couple of 3 = 8. 33 s versus? u, capital t u sama dengan v? in = being unfaithful? (? 2)? 3 = 15 meters s”1 “1 (c) By a = Average speed twenty = 8. 33 sama dengan 2 . 4 m s”1 Her normal speed for the entire trip is 2 . meters s”1. The original velocity of the skater is usually 15 meters s. almost 8 (a) The object initially moves towards the remaining and increases towards the remaining. It will speed up. 6 six 8 being unfaithful 10 C C C B A Magnitude of displacement sama dengan 2000 a couple of + 6000 2 sama dengan 6324. 6 m Value of average velocity 6324. 6 = 4? 3600 = zero. 439 meters s”1 6000 tan? = 2000? = 71. 6 His average velocity is usually 0. 439 m s”1 (S 71. 6 E). (b) The thing initially goes towards the proper and boosts towards the kept. It will reduce. Its velocity will be actually zero and then boosts in the adverse direction (moves towards the left). Revision exercise 1 Qmc (question multiple choice ) (p. 5) 1 a couple of 3 C D N eleven C Total time = 13 minutes = 780 s 840? 2 sama dengan 2 . 15 m h? 1 Typical speed = 780 (b) Displacement via Sheung Shui to Elektrotriebfahrzeug Ma Chau 1000 sama dengan? 6. a few 1 = 6300 m Magnitude of average speed 6300 = 359 = 17. your five m s”1 (1M) (1A) (1M) (1A) 12 13 D (HKCEE 2003 Newspaper II Q3) Conventional (p. 37) one particular Total period left for the two players = 5? 60 & 9 + 5? 70 + of sixteen = 565 s Total time they have been playing = 2? 70? 60? 565 = 6635 s (= 110 min 35 s = one particular h 50 min thirty-five s) (1A) 5 (a) Total distance = 1500 + forty? 1000 & 10? multitude of = 51 500 meters Total period = a couple of? 3600 & 3? 62 + 8 = 7388 s Normal speed 51 500 = 7388 = 6. six m s”1 (1M) (1A) 2 (a) 50 meters (1A) (b) Magnitude of average velocity of Kitty 50 = (1M) 1? 60 + 15 = 0. 667 m h? 1 (1A) (1M) (1A) (c) Average speed of the coach 5 + 55 + a few = 1? 60 & 15 = 0. 8 m h? 1 (b) Swimming: Average speed 1500 = 21? 60 + 28 = 1 . 16 m s”1 Cycling: Typical speed forty five 000 = 1? 3600 + one particular? 60 + 53 = 10. almost eight m s”1 Running: Normal speed 10 000 sama dengan 39? 60 + forty seven = 5. 19 m s”1 (1M) His normal speed was the highest in cycling. (1A) 3 (a) Since the lady measures the time interval based on 1 pattern of the pendulum, the error (0. three or more s) in measuring the cycle of the pendulum builds up. is from 8 to 14 s. 1A) (1A) The range of times interval (10 cycles) (b) When finding the time for one pendulum cycle, Jenny should period more pendulum cycles (e. g. 20) with the stop-watch and divide the time by the number of periods. (1A) four (a) Time required several. 4? multitude of = twenty. 6 = 359 s (5 minutes 59 s) (1M) (1A) (c) Yes. Considering that the time time period of this competition is quite extended, (1A) applying stop-watch will not likely result in large percentage problem as the reaction time for an average person is only 0. a couple of s. (1A) (1M) (c) Total time = five min forty-five s? 1 min fifty eight s = 3 minutes 47 s i9000 = a few? 60 & 47 = 227 s i9000 v? u a= (1M) t 431? 0 = 3. sama dengan 0. 527 m s”2 (1A) 227 The average speeding of the train is zero. 527 meters s”2. 6th (a) sixth is v = u + by =0+6? some = twenty-four m s”1 = eighty six. 4 kilometers h eighty six. 4 kilometers h. “1 “1 (1A) The maximum rate of the car is almost eight (1M) (a) Total length = 8000 + four thousand + 5000 = 17 000 m Total period = one particular? 3600 & 30? 60 + 45? 60 (b) v sama dengan u & at sama dengan 24 + (“4)? a couple of = of sixteen m h “1 “1 = 57. 6 km h (1A) “1 = 8100 h Average velocity 17 500 = 8100 = installment payments on your 10 m s”1 (1M) (1A) (c) The final velocity of the car is 57. 6 km h. v? u a= (1M) big t 16? 0 = 6 = installment payments on your 67 m s”2 2 . 67 meters s”2. (1A) The average speeding of the car is (b) 7 (a) Average velocity 30 1000 = 8? 60 sama dengan 62. m s”1 The standard speed from the train is usually 62. five m s”1. (1M) (1A) (b) Optimum speed 430 = sama dengan 119. 5 m h? 1 &gt, average rate 3. 6 (1A) The average speed must be smaller than the ideal speed because the train must speed up via start and slows down to stop during the trip. (1A) Magnitude of displacement = 3 thousands 2 + 4000 a couple of = 5000 m Degree of normal velocity 5000 = sama dengan 0. 617 m s”1 8100 four thousand tan? sama dengan 3000 (1A)? = 53. 1 His average velocity is zero. 617 m s (N 53. 1 E). “1 (1A) 9 (a) Distance went = 10. 5? several? 60 = 1890 m (1M) (1A) 10 (a) Total range = (120 + 50)? 1000 sama dengan 170 1000 m (1M) (1A) b) Circumference from the track =2 r = 2 (400) = 2513 m The length travelled by simply Marilyn is 3 1890 m which can be about of the 4 circumference. (1A) (b) N? XYZ is a right-angled triangle. Z .? 50 kilometers 30 Con 60 Times? 120 km Magnitude of displacement (from town X to community Z) sama dengan 120 500 2 & 50 500 2 sama dengan 130 500 m 120 tan? sama dengan 50? = 67. 4 Magnitude of displacement AB = 4 hundred 2 & 400 2 (1A) (1A)? = 90? 67. 4 = twenty-two. 6? sama dengan 60? twenty two. 6 = 37. 4 The total shift of the car is 130 000 meters (N thirty seven. 4 E). = 566 m Size of average velocity 566 = several? 60 sama dengan 3. 13 m t 400 color? = 4 hundred? = 45 (S 45 E). “1 (c) (1A)

Total time 170 1000 = sama dengan 10 two hundred s 62 3. six Magnitude of average speed 130 500 = 10 200 = 12. six m s”1 Its typical velocity is 12. six m h (N thirty seven. 4 E). “1 (1A) (1A) (1M) (1A) Her average speed is several. 14 meters s”1 11 (a) AC = 60 2 + 70 2 sama dengan 100 meters 80 color? =? = 53. 1 60 (1M) The total shift of the sportsperson is 100 m (S53. 1W). (1A) 13 (Correct label of velocity with correct path (towards the left). ) (Correct ingredients label of speeding with correct direction (towards the right). ) (1A) (1A) (a) The coin moves inside the following series: B A C C A Consequently , it is for A finally. Displacement with the coin sama dengan 15 centimeter (1A) (1M) (1A) (1M) b) Length travelled by the coin sama dengan 15 & 30 + 30 sama dengan 75 centimeter (b) Period / s i9000 v as well as m s”1 0 “6 1 “4 2 “2 3 0 4 +2 5 +4 6 +6 (1A) (1A) (c) (i) Total period = two s? 5 = almost eight s Average velocity 15? 10? a couple of = almost 8 = 0. 0188 meters s? you (0. 5A? 6) (1M) (1A) (c) The car will certainly slow down as well as its speed is going to drop to zero. There after the car will move towards the right with increasing speed (uniform acceleration). (1A) (1M) (1A) (1M) (1A) (1M) (1A) A (ii) Average speed seventy five? 10? a couple of = 8 = zero. 0938 m s? you (1M) (1A) 12 (a) Total range travelled sama dengan 60 & 80 & 80 + 60 sama dengan 280 m (d) (i) The gold coin moves inside the following series: B A C C A N B b) Magnitude of total shift = 80 + 70 = one hundred sixty m one hundred sixty m (west). The total displacement of the sportsman is Therefore , it is in B finally. zero. the coin is likewise zero. (1A) (1M) (1A) (1M) (1A) (1M) (1A) (ii) The displacement of the coin can be Therefore the normal velocity of (c) Total distance journeyed = 280 + 60 + eighty = four twenty m 16 (a) Total distance sama dengan? r sama dengan 5? 60 m C = 15. 7 m Total displacement =5+5 = 10 meters 80 m The overall displacement went by her is 10 m. (b) Jane’s affirmation is inappropriate. (1A) Seeing that both young ladies start at Times and meet up with at Sumado a, they have a similar displacement. (1A) Betty’s declaration is wrong. 1A) Since both girls return to their starting point, their particular displacements are zero. (1A) Physics in articles (p. 40) (a) From 19 January 06\ to 28 February 2007, (1A) It takes New Horizons spacecraft a total of 406 times to travel from the Earth to Jupiter. (1A) (b) (i) Average rate total length travelled sama dengan total time of travel (1M) = almost 8? 108 406? 24 (1A) (1M) = 8. twenty-one? 104 kilometers h? 1 (ii) Normal acceleration difference in velocity = total time of travel = (8. 3? 5. 79)? 10 four 406? twenty four = 2 . 50? 104 km l? 2 (1A) (1A) (c) July 2015 2 1 a couple of 3 some 5 Action II 10 (a) The item moves with a constant elocity. Practice installment payments on your 1 (p. 61) Deb B Deb D W 30? 12 = twelve m s”1 v= two (b) The thing moves having a uniform speed from others. (c) The object moves having a uniform deceleration, starting with some initial velocity. Its speed becomes absolutely no finally. The speed of the car at t = 2 s can be 10 meters s”1. six 7 C (d) The thing first goes with a standard acceleration from rest, in that case at a continuing velocity, and finally moves using a smaller uniform acceleration again. (a) Total displacement sama dengan 4? five + (? 5)? (7? 5) sama dengan 10 m The total shift from the staircase to her class is 10 m. (e)

The object techniques at a constant velocity then suddenly moves at frequent velocity of same degree in the opposing direction. (b) Classroom C 8 (f) The object moves with homogeneous deceleration via an initial velocity to rest, and continue to move with the homogeneous acceleration of the same magnitude in opposite direction. 9 (a) The object boosts. (b) The item first techniques with a frequent velocity. Then it becomes standing and finally moves with a bigger constant speed again. eleven (a) The object moves with zero acceleration (with frequent velocity of 50 m s”1). (b) The item moves which has a uniform cceleration of five m s”2. (c) 12 The object techniques with homogeneous deceleration of 5 meters s”2. (c) The object decelerates to rest, and after that accelerates in opposite way to return to it is starting point. (a) It goes away from the sensor. (d) The object moves with uniform velocity towards the origins (the absolutely no displacement position), passes the origin, and continue to be move away from origin while using same homogeneous velocity. (b) (c) The greatest charge of enhancements made on speed zero? 3. 5 = two = “1. 75 m s”2 (d) Total length travelled = area beneath the graph several. 5? a couple of 2? six = + 2 a couple of = being unfaithful. 5 m Practice installment payments on your 2 (p. 71) one particular

C By simply v2 sama dengan u2 & 2as, 290 3. 6th 2 13 (a) =0+2? 1? s i9000 s = 3240 meters = a few. 24 kilometers &lt, several. 5 kilometres The minimum length of the catwalk is 3. 5 kilometers. 2 M Cyclist Back button is going at continuous speed. Time for cyclist Back button to reach complete line shift 150 = = = 30 s i9000 time a few For cyclist Y: u = a few m s”1, s sama dengan 250 meters, (b) Total distance went = place under the graph (12 + 6)? several = two = 27 m a = a couple of m s”2 By t = lace + 1 2 in, 2 you 250 sama dengan 5? t +? two? t2 two (c) Normal speed total distance journeyed = time taken twenty-seven = three or more t = 13. 5 s or perhaps t sama dengan? 18. a few s (rejected) Y requires 13. a few s to succeed in finish range. Therefore , bicyclist Y will certainly win the race. 3

B Because the bullet start decelerates after fired in the wall, we’re able to just consider the shift of the topic in the wall structure. To prevent the bullet coming from penetrating the wall, the bullet must stop in the wall. sama dengan 9 m s”1 13 (a) The lady moves on the motion sensor. (b) The highest speed in the girl inside the journey is definitely 3. 5 m s”1. By simply v2 sama dengan u2 + 2as, zero = 500 + two? (? 800 000)? s i9000 2 8 By versus = u + for, 14 = u & 2? five u sama dengan 4 m s”1 s = zero. 156 meters = 15. 6 cm &lt, 15. 8 cm The bare minimum thickness with the wall is 15. 8 m. Simply by v2 = u2 & 2as, 142 = forty two + 2? 2? t s = 45 meters 4 C When the puppy catches the thief at t = 5 s, its total displacement can be 30 m.

The dog is sitting primarily, so u = zero. 1 By simply s sama dengan ut + at2, a couple of 1 40 = 0 + a(5)2 2 The displacement of the girl is usually 45 meters. 9 (a) v sama dengan u + at sama dengan 0 + 20? 0. 3 = 6 m s? one particular The horizontal speed in the ball travelling towards the goalkeeper is 6 m t? 1 . a = 2 . 4 m s”2 Its acceleration is definitely 2 . some m s”2. (b) Simply by v2 sama dengan u2 + 2as, 02? 62 a= = “22. 5 m s? two 2? 0. 8 The deceleration with the football needs to be 22. your five m s i9000? 2 . five 6 M 90 thirty six? v? u = a few. 6 3. 6 sama dengan 1 . your five m s”2 a= t 10 By simply v sama dengan u & 2as, 2 2 12 (a) The response time of the cyclist can be 0. 5 s. s= v? u = 2a 2 2 90 three or more. 6 thirty eight 3. 6th 2? 1 . 5? 2 2 = 175 m (b) Braking system distance (2.? 0. 5)? 15 = 11. twenty-five m = 2 Thinking distance sama dengan 15? zero. 5 sama dengan 7. five m Halting distance sama dengan 11. 25 + several. 5 = 18. seventy five m kid. 20 m The distance travelled by the motorbike is 175 m and its particular acceleration is 1 . your five m h. “2 7 (a) Pondering distance = speed? effect time 108 =? 0. 8 sama dengan 24 meters 3. six Therefore , the bicycle will not hit the (b) Because the car decreases uniformly, stopping distance v+u =? to 2 108 +0 = 3. six? (3? zero. 8) 2 = thirty-three m 14 By versus = u2 + 2as, 0 = 32 + 2? (“0. 5)? h s=9m 8m Therefore , the golf ball can reach the hole. 2 12 (a) (i) By sixth is v = u + in, 0 = u + (“4)(4. 75) u sama dengan 19 meters s”1

The original velocity of the car is definitely 19 m s”1. (c) Stopping length = thinking distance & braking distance = twenty-four + 33 = 57 m (ii) Simply by v2 = u2 + 2as, zero = nineteen + 2? (“4)? s s = 45. 1 m a couple of 3 C For choice A, apply equation v2 = u2 ” 2gs and take s sama dengan 0 (the ball results to the second floor), v = “u = “10 m s”1 (vertically downwards) The shift of the car before that stops ahead of the traffic light is 45. 1 meters. This is the same velocity while the initial speed of option B. Consequently , in both ways the ball has the same straight speed mainly because it reaches the ground. (b) By v sama dengan u + 2as, 17 = zero + two? 3? h s = 48. two m 2 2 2

The displacement of the car between beginning from rest and moving for 17 m s can be 48. a couple of m. “1 4 B Take the upward direction because positive. 1 By s = ut + at2, 2 one particular 0 = u? 31 +? (? 10)? 302 2 u = 150 m s”1 13 (a) By a huge selection of = u2 + 2as, v2 = 0 + 2? zero. 1? 500 v sama dengan 10 meters s”1 His speed is 10 meters s. “1 (b) Consider the initial section. Simply by v = u + at, v? u t= a 10? 0 = 0. 1 = 100 s Consider the 2nd section. you By t = ut + at2, 2 1 700 = 10t +? 0. 5t2 a couple of t = 40 s i9000 or big t = “80 s (rejected) The speed of the bullet is 150 m s”1 if it is fired. 5 Speed of stone Equation used t=1s t=2s t=3s t=4s sixth is v = u + for Distance travelled by the stone 1 s = ut + in 2 two m 20 m 45 m eighty m 10 m s”1 20 meters s 40 m s i9000 “1 “1 40 meters s”1 Total time taken = 100 + 40 = 150 s It will require 140 s i9000 for Jason to travel all downhill. 6 1 By t = ut + at2, 2 1 10 sama dengan 0 + (10) t2 2 to = 1 ) 41 s i9000 v = u + at Practice 2 . 3 (p. 83) 1 2 D Deb = 0 + 10(1. 41) sama dengan 14. one particular m s”1 It takes 1 ) 41 h for a diver to drop by a 10-m platform. His speed is definitely 14. one particular m s”1 when he makes its way into the water. 7 Take the upward course as confident. By v = u + 2as, 4 sama dengan 0 + (2)(“10)s s = 0. 8 m 2 two 2 Besides, since Con spends a shorter time to reach the highest stage, it should be dismissed after X. 10 (a) By t = ut + The best position come to by the puppy is 0. m above the ground. almost 8 (a) Consider the kid’s downward trip. Take the downwards direction because positive. you By s i9000 = ut + at2, 2 one particular 0. 5 = zero + (10) t2 a couple of t = 0. 316 s you 2 for, 2 1 120 = 8t +? 10? t2 2 capital t = some. 16 s i9000 or t =? five. 76 s (rejected) It will require 4. sixteen s to reach the ground. (b) v sama dengan u & at = 8 + 10? four. 16 = 49. 6 m s”1 Its velocity on striking the ground is definitely 49. 6th m s”1. 11 (a) Distance between ceiling and her hands = 6th ” 2 ” 1 ) 2 sama dengan 2 . almost 8 m Hang-time of the youngster = 0. 316? 2 = 0. 632 s (b) Permit s always be her top to bottom displacement the moment she gets. As the most jumping speed is eight m s”1, i. e. u = 8 m s”1. By simply v2 sama dengan u2 & 2as, v2? 2 s= 2a a couple of 0? 82 = (upwards is positive) 2? (? 10) s i9000 = 3. 2 m &gt, 2 . 8 m Therefore , the indoor recreation space is unsafe for playing trampoline. one particular (a) By simply s sama dengan ut + at2, a couple of 1 132 = 0? t +? 10? t2 2 to = 5. 14 t The vehicle can experience a free fall in the Zero-G center for five. 14 h. (b) Take the upward way as great. By versus = u + 2as, 0 = u & 2? (“10)? 0. five u sama dengan 3. sixteen m s”1 2 a couple of 2 The jumping rate of the young man is three or more. 16 meters s”1. 9 Take the upward direction since positive. (a) By v2 = u2 + 2as, 0 sama dengan u2 & 2(“10)(200) u = 63. 2 m s”1 The speed of the firework X is usually 63. a couple of m s”1 when it is dismissed. 12 (b) By sixth is v = u + at, = 63. 2 & (“10)t capital t = 6th. 32 h It takes six. 32 t for the firework By to reach that height. (c) From (a) and (b), for firework Y to blow up at 140 m over a ground, the speed of Y should be less space-consuming than that of By. Therefore , Con should be dismissed at a (b) Simply by v2 = u2 & 2as, v2 = 02 + 2? 10? 132 v sama dengan 51. four m t? 1 The velocity of the motor vehicle before considering a stop can be 51. 5 m h? 1 . lower rate. (c) Take the upward course as confident. By sixth is v = u + for, “v = v ” gt 2v = grand touring If the natural stone is forecasted with a velocity of 2v, let the fresh time of travel around be t?. (“2v) sama dengan (2v) ” gt? versus t? sama dengan 4 ( ) g = 2t Its new time of travel is 2t. 6

M Take the up direction because positive. you s sama dengan ut + at2 a couple of 1 = (10)(4) + (“10)(4)2 2 = “40 m The distance between the sandbag and the earth is forty five m in order to leaves the balloon. Revision exercise two Multiple-choice (p. 87) one particular D By v2 sama dengan u2 & 2as, 0 = 102 + 2a(25 ” 12? 0. 2) a sama dengan “2. seventeen m s”2 His minimum deceleration is definitely 2 . 18 m s”2. 2 a few D W Consider the rock released from the 2nd floor. Simply by v2 sama dengan u2 & 2as, v2 = 2as floor. Remember that s2 = 3. 5s. (v2)2 sama dengan 2as2 sama dengan 3. 5(2as) = 3. 5v2 a huge selection of = 1 . 87v (as u sama dengan 0) Then consider the rock released from the 7th 7 8 D C Take the down direction since positive. u = 200 m s”1, v sama dengan 5 meters s”1, a =? 0 m s”2 By sixth is v = u + for, 5 sama dengan 200 + (? 20)t t sama dengan 9. 75 s The rockets needs to be fired no less than 9. seventy five s. The two C and D gratify this need. But for G, after firing for twelve. 2 t, v = u & at = 200 + (“20)(10. 2) = “4 m s”1 i. electronic. it lures away from the Moon with four m s”1 upwards. It cannot arrive at the Celestial body overhead. Therefore , the best answer is C. four 5 A C The stone returns to the surface with the same speed (but in opposite direction). being unfaithful 10 G D 11 12 13 (HKCEE 2006 Newspaper II Q1) (HKCEE 3 years ago Paper 2 Q2) (HKCEE 2007 Conventional paper II Q33) (b) (i) Conventional (p. 89) 1 (a) The reaction time of the driving force is 0. 6 s. (b) sixth is v a= to = 0? 12 a few. 6?. six (1A) (Correct axes with label) from t sama dengan 1 . 20 s to 1. 25 s) from capital t = 1 ) 45 h to 1. 55 s) (1A) (1A) (1A) (A straight line with slope sama dengan 0. thirty five m s”1 (A direct line with slope sama dengan “0. thirty five m s”1 (1A) (1M) = “4 m s”2 The speeding of the car is “4 m s”2. (c) The stopping length of the car is the place under chart. Stopping length 12? (3. 6? 0. 6) =12? 0. 6 + 2 = twenty-five. 2 m The blocking distance in the car is usually shorter than 27 m. The driver will never be charged with driving previous a reddish light. (1A) (1A) (1M) (ii) two (a) The item moves away from motion sensor with uniform velocity by 0. thirty-five m s”1 from to = 1 ) 20 s i9000 to 1. twenty-five s. 1A) From to = 1 . 25 s i9000 to 1. 45 s, the thing moves with negative speed. (1A) Then, from to = 1 . 45 t to 1. 40 s, the thing changes their moving direction and techniques towards the movement sensor again with a consistent velocity of “0. thirty five m s”1. (1A) (Correct axes with labels) (1A) (Correct graph with the speed of? 0. 35? zero. 35 regarding 1 . forty five? 1 . 35 = “7 m s”2 at capital t = 1 ) 30 s to 1. 45 s) (1A)! a few (a) (b) Total shift of the car = location bound by the v? big t graph and the time axis 1 you = (5? 5)? (20? 3) a couple of 2 =? 17. your five m (1M) (1A) (c) Yes, the auto moves 12. 5 m forwards from t sama dengan 0 to t = 5 h. Therefore , it hits the roadblock. 1A) 5 Take the upward way as great. (a) From point A to the greatest point: (Correct axes with labels) (Correct shape of minibus’ graph) (Correct shape of sports car’s graph) (Correct values) (1A) (1A) (1A) (1A) By a huge selection of = u2 + 2as, 0 sama dengan 42 + 2 (“10) s s = 0. 8 meters By v = u + in, 0 sama dengan 4 + (“10)t big t = 0. 4 h (1M) From your highest point out the playground equipment: 1 h = ut + at2 (1M) a couple of 1 = 0 + (“10)(1. 2 ” zero. 4)2 2 = “3. 2 m (1A) several. 2 m above the playground equipment. (1A) The most height come to by him is (1M) (b) Through the graph in (a), the two vehicles have similar velocity at t? installment payments on your 3 t after moving the traffic light. (1A) (1M) (c)

The area underneath graph is the displacement from the cars. Consider their displacements at big t = several s, To get the expensive car: 1 h =? 15? 3 = 22. 5 m a couple of For the minibus: you s sama dengan? (7 & 13)? three or more = 40 m two The minibus will take the lead a few s following passing the traffic light. (1A) (b) Height of point A above the trampoline (1A) sama dengan 3. two ” 0. 8 sama dengan 2 . 4 m (1M) (1A) 6 (a) Primary velocity versus = 90 km h”1 90 sama dengan m s”1 3. 6 = 25 m s”1 Thinking range =v? t = twenty-five? 0. a couple of =5m The thinking length is 5 m. (1A) (1M) some (a) The automobile moves ahead with consistent acceleration by? 1 meters s? a couple of from big t = 0s to t = five s. (1A) (1A) Then the car changes its shifting direction.

Coming from t = 5 s i9000 to to = eight s, it moves back with a uniform acceleration of? 6. 67 m t.? 2 The instantaneous speed is 0 at big t = a few s. (1A) ” (b) By v2 sama dengan u2 + 2as, a huge selection of? u2 a= 2s two 0? 25 2 = 2? (80? 5) sama dengan? 4. seventeen m s”2 4. seventeen m s”2. (1M) (c) The slope of the chart is the value of the speed of the apple. speed as well as m s? 1 several. 75 (1A) (1A) Consequently, the deceleration of the car is (c) By v2 = u2 + 2as, s= versus? u 2a 0 a couple of? 25 2 = a couple of? (? 5. 17? 2) 2 two (1M) 0 0. 775 time as well as s (Correct labelled axes) (2A) (1A) (Straight series with a slope of twelve m h? 2) = 37. a few m Braking system distance sama dengan 37. a few m Blocking distance sama dengan 37. 5 + a few = 40. m (1M) (d) Both graphs have no difference. (1A) (1A) almost 8 (a) Take the downward path as great. By v2 = u2 + 2gs, v sama dengan u + 2 gs 2 The driver could not stop before the targeted traffic light. Consequently , his state is incorrect. (1A) (1M) 7 (a) Take the downward direction as positive. 1 By s = lace + gt2, 2 you 3 sama dengan 0? capital t +? 15? t2 a couple of 3? 2 t= sama dengan 0. 775 s 10 (1M) = 0 2 + a couple of? 10? (40? 3) = 27. 2 m s”1 cushion is 27. two m s i9000? 1 . one particular (b) (i) By s = lace + gt2, 2 1 40 ” 3 = 0 &? 10? t2 2 capital t = 2 . 72 s (1A) The speed of the residents landing within the (1M) (1A) The apple travels in air pertaining to 0. 775 s. (1A) (b) Simply by v2 sama dengan u2 + 2as, versus = a couple of? 10? three or more (1M) 1A) “1 sama dengan 7. seventy five m s? 1 The speed of the apple is six. 75 meters s if the apple simply reaches the land. The time of travel in air is usually 2 . 72 s. u+v (ii) Simply by s = t, (1M) 2 2s t= u+v 2? several = capital t 27. 2 + 0 = zero. 221 s i9000 (1A) Enough time of get in touch with is zero. 221 t. (c) (b) Slope of the graph from big t = 0 to t = zero. 28 s i9000 2 . three or more? 0 = 0. 28? 0 sama dengan 8. twenty one m s”2 The speeding of the ball due to gravity is eight. 21 m s”2. (1M) (1A) (c) (Correct branded axes) (Correct shape) (Correct values) (1A) (1A) (1A) (i) 9 (a) capital t = two s: Displacement of the trolley = zero. 7? zero. 15 = 0. fifty-five m to = three or more. 4 h: (1A) Displacement of the cart = 1 . 175? 0. 15 = 1 . 025 m capital t = some. 9 h: 1A) Displacement of the cart = 0. 6? zero. 15 sama dengan 0. forty five m (1A) (b) That moves away from the motion messfühler with a changing speed via t = 2 h to big t = 3. 4 s i9000. (Correct sign) (Correct shape) (1A) (1A) (1A) (1A) (1A) (ii) The method can not work Then it rests momentarily for t sama dengan 3. 4 s. There after, it goes towards the movement since ultrasound will be shown by the clear plastic platter. (1A) (c) sensor using a changing acceleration. 1 Simply by s sama dengan ut + at2, two 1? zero. 1 = 0. 7? 2 . being unfaithful +? a? (2. 9)2 2 a =? 0. 507 m s? 2 (1A) (1M) 11 (a) (i) The ball is usually held 0. 15 meters from messfühler before released. The ball hits the ground which is 1 ) m through the sensor. (1A) (1A) Consequently , the ball drops a height of 0. 96 m. that are 0. forty five m, 0. 65 meters and 0. 775 meters from the sensor in its initial 3 springs back. (1A) The acceleration in the trolley is definitely? 0. 507 m t? 2 . (ii) The ball rebounds for the positions twelve (a) The motion sensor is protruded outside the stand to avoid the reflection of ultrasonic signal from desk. (1A) At the initial rebound, the ball rises up (1. 1? zero. 45) sama dengan 0. 65 m. nd The average velocity is 66. 6 meters s”2. (1A) (1A) (1A) (c) v / m s? you 6. thirty-two At the two rebound, the ball goes up up (1. 1? zero. 65) sama dengan 0. 45 m. rd At the three or more rebound, the ball increases up (1. 1? zero. 75) = 0. 325 m. (b) (i) The ball visitors the ground with velocities of 3. 9 meters s, 3. 25 m s and 2 . seventy five m s”1 in its first 3 springs back. (3A) three or more. 9 (1M) 0. ninety five? 0. fifty-five (1A) “1 “1 t3 t1 t2 t4 t5 t/s (ii) Acceleration = slope of graph = = 9. 75 m s”2? six. 32 (3 straight lines) (Correct slopes) (1A) (1A) 12 Take the downward direction as positive. 1 (a) By s i9000 = ut + gt2, (1M) two 1 a couple of = 0? t +? 10? t2 2 a couple of? 2 t= = 0. 632 h (1A) 12 It takes 0. 632 s from t1 to t2. (Correct brands of time and velocity)(1A) 13 (a) Acceleration v = 70 km h”1 70 = meters s”1 three or more. 6 sama dengan 19. 5 m s”1 d Reaction time = v 6 = nineteen. 4 sama dengan 0. 309 s The response time of the man was 0. 09 s i9000. (1M) (b) At t2, v sama dengan u & at (1A) = 0 + twelve? 0. 632 = six. 32 m s “1 “1 (1M) Shirley’s acceleration is 6. 32 m s once she lands on the playground equipment at t2. At t4, she leaves the trampoline at the same acceleration. Therefore , coming from t3 to t4, simply by v2 sama dengan u2 + 2as, a= v2? u2 2s (? 6. 32) 2? 0 2 = 2? 0. 3 (b) By v2 = u2 + 2as, v2? u2 a= 2s 2 zero? 19. 5 2 = 2? twenty four = “3. 92 m s”2 several. 92 m s”2. (1M) (1M) (1A) The average deceleration of the car was (c) (1A) Speed v = 80 km h”1 eighty = meters s”1 three or more. 6 = 22. a couple of m s”1 = sixty six. 6 meters s”2 Thinking length = vt = twenty two. 2? zero. 309 sama dengan 6. 86 m By simply v sama dengan u & 2as, braking distance s i9000 v2? u2 = 2a 2 zero? 22. 2 2 = 2? several. 92) 2 2 (1A) Take the up direction since positive. you s = ut + at2 (1M) 2 one particular = 7? 1 . seventy five +? (“10)? 1 . 745 2 = “3. summer m (negative means this particular is below the spring board) The planting season board is usually 3. summer m over a water. Substitute method: (1A) = sixty two. 9 meters Therefore , the stopping range = 6th. 86 + 62. 9 = 69. 8 meters (1A) Consider the up motion and downward motion separately. For the upward motion, your woman takes 0. 7 h to reach the highest point from the spring table. Take the upward direction as positive. one particular By s = lace + at2, (1M) two 1 s1 = several? 0. 7 +? (“10)? 0. seventy two 2 sama dengan 2 . forty-five m Intended for the downward motion, the girl takes 1 ) 5 s i9000 from the top point to enter in water. Take the downward course as great. By h = ut + you 2 grand touring, 2 you s2 = 0 &? 10? 1 . 052 = 5. 51 m 2 (1A) This stopping distance is greater than the initial range between the car and the young man. (1A) Therefore , the car would have knocked down the boy in the event the car acquired travelled by 80 kilometers h? one particular or more quickly. (d) A drunk includes a longer effect time. (1A) This means that the thinking length, and thus the stopping length (sum of thinking length and braking system distance), boosts. (1A) (1M) (1A) 14 (a) Take those upward path as great. By versus = u + at, u sama dengan 0? (? 10)? 0. 7 = 7 m s”1 plank is six m h. 1 Which means height with the spring board above the normal water = s2 ” s1 = 5. 51 ” 2 . forty-five = three or more. 06 meters (1A) (1M) (1A) The speed of Belinda leaving the spring (b) Total time taken from the spring plank to the normal water = zero. 7 + 1 . 05 = 1 . 75 s (c) versus = u + by = 0 + (? 10)? 1 ) 05 =? 10. five m s”1 is 12. 5 meters s”1. The speed from the diver going into the water (d) Deceleration of car Y = incline of the chart during zero. 5 t? 8. a few s = 0? nineteen. 4 = “2. 43 m s”2 8. a few? 0. 5 (1A) The deceleration of car Con is installment payments on your 43 meters s”2. (c) Thinking range = region under the chart during 0? 0. your five s = 19. 5? 0. 5 = on the lookout for. 7 meters (1A) (Correct shape) (Correct times) (Correct velocities) 1A) (1A) (1A) Braking range = area under the chart during zero. 5 s i9000? 8. five s you =? 19. 4? (8. 5 ” 0. 5) 2 = 77. 6th m range are 9. 7 meters and seventy seven. 6 meters respectively. (1A) The considering distance and the braking (e) (See the figure in (d). ) (Correct slope , parallel to that in (d). ) (1A) (Correct location ” above that in (d). ) (1A) 15 (a) Speed 70 km h”1 70 = m s”1 3. 6th = 19. 4 meters s “1 (d) The coloured region is equal to the difference inside the stopping miles travelled by cars X and Con. (1A) (e) (1M) Preventing distance of car By = region under the chart during 0? 5 s 1 sama dengan? 19. 5? 5 = 48. 5 m two Coloured region = on the lookout for. 7 & 77. 6th ” 48. = 32. 8 meters &lt, 60 m Because the difference in stopping miles of the cars is less space-consuming than the initial parting of the autos, the two automobiles do not wage war with each other before they quit. (1A) (1M) (1M) Length travelled by simply car Y in two s sama dengan vt sama dengan 19. 4? 2 = 38. eight m &lt, 50 meters Since the range between the automobiles is greater than the distance that car Sumado a can travelling in 2 s, the driving force of car Y obeys the guideline. corresponding v”t graph. Deceleration of car X sama dengan slope of the graph during 0? 5 s (1A) (1M) (b) Deceleration of any car is a slope of their 0? nineteen. 4 = 5? 0 = “3. 88 meters s”2 The deceleration of car By is three or more. 88 meters s”2. (1A) 16 a) From big t = 0s to big t = your five s, the automobile moves with a uniform velocity of 18? 0 = 3. 5 m s”2. 5 (1A) From t sama dengan 5 s to t = twenty s, the car moves using a constant velocity of seventeen m s”1. (1A) By t = 20 t to capital t = twenty eight s, the car moves which has a uniform speed of 0? 17 =? 2 . a hundred and twenty-five m s”2. 28? 20 at rest. (1A) (b) s = lace + 1 2 at 2 one particular = zero +? 18. 5? (8? 60)2 a couple of = two 016 1000 m (2016 km) (1M) (1A) The Shuttle trips 2 016 000 m (2016 km) in the initial 8 mins. From big t = twenty-eight s to t sama dengan 30 s, the car continues to be (1A) 19 (a) (i) The cyclist is using first items when the velocity is finest before braking system. shortest period. (1A) (1A) (1M) (1M) (1A) b) (ii) The cyclist uses second equipment for the (b) Distance travelled = area below straight range PQ (8 + 6)? 2 = 2 sama dengan 14 m The cyclist travels 18 m in second items. (c) The acceleration during t = 18 t? 20 s 0? on the lookout for = (1M) 20? 18 =? 5. 5 meters s”2 The deceleration is usually 4. your five m h. “2 (1A) (Correct shape) (Correct period instants) (Correct accelerations) (1A) (1A) (1A) (1A) (1A) 20 twenty one (c) Yes. (HKCEE 2005 Paper I actually Q1) 1 (a) h = lace + at2 2 1 = 0 +? 10? (500? twelve? 3)2 a couple of = 1 ) 25 meters Therefore the minimum height the (1M) The automobile changes path at to = 35 s. Their velocity changes from confident to negative, showing an alteration in its exploring direction. 1A) (1M) (1A) (1A) notebook must fall for it to get ‘saved’ is 1 . 25 m. (b) v = u + at sama dengan 0 & 10? (500? 10 ) = your five m h? 1 the land is five m s”1.? 3 (1M) (1A) seventeen 18 (HKCEE 2002 Newspaper I Q8) (a) v = u + in = 0 + seventeen. 5? almost 8? 60 sama dengan 8400 m s”1 mins is 8400 m s”1. The speed with the computer when it hits The velocity of the Shuttle service after the initially 8 (c) Many falls could be from below this elevation, effect. (1A) (1A) (1A) so the safeguard will not have considered Physics in articles (p. 96) (a) 2 . forty-five m (b) (i) By simply v2 sama dengan u2 + 2as, u = sixth is v? 2as u2 = zero? 2(? 10)(2. 45 + 0. ’07? 1 . 09) u = 5. thirty-five m t? 1 2 2 (1A) (1M)

Take the upward course as positive. 22 (a) Any one via: Rate of change of displacement Shift per unit time (1A) (b) The velocity of a braking car can be decreasing (with time) (1A) so the car has adverse acceleration. (1A) Its shift is (still) increasing over time, so it is velocity can be (still) great In this case, the acceleration and velocity happen to be in opposite directions. (1A) (1A) (1A) The vertical speed of Javier Sotomayor is 5. 35 meters s? you when he leaves the ground. (ii) Take the up direction because positive. Consider the upward journey. By v = u + at, sixth is v? u zero? 5. thirty-five t= sama dengan = 0. 54 s a? 10 (1M) (c) i) Consider the downward journey. 1 By t = ut + at2, (1M) a couple of 1? (2. 45 + 0. 07? 0. 71) = 0 + (? 10) t2 2 big t = zero. 60 s The time that he stays on in the air sama dengan (0. fifty four + 0. 60) sama dengan 1 . 14 s Substitute method: (1A) (Correct graph) (1A) Take those upward direction as confident. 1 By simply s = ut + at2, (1M) 2 (0. 71? 1 . 09) sama dengan 5. 35t + one particular (? 10)t 2 (1M) 2 capital t = 1 . 14 t or big t =? zero. 07 s i9000 (rejected) (ii) Vertical length travelled sama dengan area underneath the graph from 4. 0s to 15. 0 s (70 + 130)? 6 sama dengan 2 (1M) (1A) Enough time that he stays up is 1 ) 14 s. = six-hundred m (1A) The straight distance journeyed by the explode between capital t = some. 0 s and t sama dengan 10. h is 600 m. 3 1 2 a few 4 C C Force and Action 6 (a) The MTR train is usually accelerating inside the forward way. The man has a tendency to move at his initial speed (smaller speed), thus he would maneuver backwards in accordance with the MTR train. (b) The MTR train is slowing down. The man tends to push at his original rate (greater speed), so he would move ahead relative to the MTR coach. (c) The MTR educate is shifting forwards in constant speed. The man moves forwards with the same regular velocity, and so he would stay at rest in accordance with the MTR train. (d) The MTR train can be turning a corner. The Practice 3. you (p. 104) (b), (e), (f) five a) Stretching out a rubber band (b) Standing on the ground (c) Going for walks time (e) (f) A compass A rubbed plastic-type ruler allures small bits of paper (d) Exists in every single object on the earth at any 7 gentleman tends to approach at his original path, so he’d move outwards relative to the MTR coach. In space, the gravitational force works on the spaceship is negligible. When the rockets are turn off, they do not apply a push on the spaceship. Therefore , simply no net power acts on the spaceship. By simply Newton’s 1st law, the spaceship is within uniform action and can travelling far out in space. 8 Joan moves on the ice surface area with a continuous velocity.

Practice 3. 2 (p. 111) 1 2 3 4 5 C C Deb C (a) No . Sports athletes would hit the wall of the arena if it is also close to the polishing off line. (b) The pad is used to protect the sports athletes if that they hit the wall after passing the finishing range. Practice 3. 3 (p. 122) you 2 a few 4 five D A B A D 6 (a) 7 (a) Horizontal component = forty five + 40 cos 30 = 66. 0 N Vertical aspect = 40 sin 30 = 15 N Resultant = sixty six 2 & 15 two = 67. 7 In Let? always be the viewpoint between the resulting Resultant’s degree is 67 N plus the angle between resultant as well as the horizontal is usually 13. (b) and the lateral. 15 tan =? sama dengan 12. 8 66 Resultant’s magnitude is usually 67. And and the viewpoint between the resultant and the side to side is doze. 8. (b) Horizontal element = 40 + 30 cos 45 = sixty one. 2 D Vertical part = 40 sin 45 = 21 years old. 2 N Resultant’s magnitude is 66 N plus the angle between your resultant plus the horizontal is usually 19. (c) Resultant sama dengan 61. a couple of 2 + 21. a couple of 2 = 64. eight N Allow? be the angle involving the resultant and the horizontal. 21 years old. 2 tan =? = 19. 1 61. 2 Resultant’s magnitude is sixty four. 8 N and the perspective between the resultant and the horizontal is nineteen. 1. (c) Resultant’s value is 60 N plus the angle between the resultant and the horizontal is usually 25. (d)

Horizontal element = forty + 31 cos 60 = fifty-five N Top to bottom component sama dengan 30 bad thing 60 sama dengan 26. zero N Resultant = 55 2 + 26. zero 2 = 60. eight N Permit? be the angle between resultant as well as the horizontal. 26. 0? sama dengan 25. 3 tan sama dengan 55 Resultant’s magnitude can be 60. almost 8 N as well as the angle between the resultant plus the Resultant’s magnitude is 55 N as well as the angle between your resultant plus the horizontal is 37. side to side is 25. 3. (d) Resulting = 40 2 + 30 2 = 55 N Allow? be the angle between resultant as well as the horizontal. 40 tan sama dengan? = thirty eight. 9 forty five Resultant’s magnitude is 50 N and the angle involving the resultant plus the horizontal is 36. 9.

Hence, the angle between two 5-N forces is 120. Option method: By simply tip-to-tail technique, the two 5-N forces as well as the resultant 5-N force type an equilateral triangle. It truly is known that every angle associated with an equilateral triangular is 60. Therefore , the angle between the two 5-N forces is 120. eight (a) 10 (b) Resulting force = 2? four hundred = 800 N The resultant push provided by the cable can be 800 D. 11 Pertaining to the 2-kg mass: (c) 9 Ur = pounds? cos? sama dengan 20 cos 30 = 17. a few N Suppose the two pushes act inside the direction while shown. To = twenty N As a result we have: Straight component Forex = five sin? Horizontally component Fy = five? 5 cos? = 5? 1? cos? ) (magnitude of the resultant)2 = Fx2 + Fy 2 52 = (5 sin? )2 + [5? (1? cos? )]2 1 = trouble? + you? 2 cos? + cos? 2 2 2T cos 45 sama dengan W 2? 20? cos 45 = W cos? = zero. 5 T = twenty-eight. 3 D? = 60 12 (a) 2T sin 10 = 500 T sama dengan 1440 N The tension from the string is 1440 And. 3 4 5 6 B C A Net force = ma sama dengan 40? 0. 5 sama dengan 20 And C By v2 ” u2 sama dengan 2as, zero ” u2 = 2a(20)? u2 sama dengan 40a u2 a=? 45 Resistance sama dengan ma sama dengan 12? u2 = “0. 03u2 40 (b) Component of force = T cos 10 = 1440? cos 10 = 1420 And The component of the power that drags the car is definitely 1420 And. 13 (a) 7 8 ‘A bag of glucose weighs twelve N. ‘ or ‘A bag of sugar contains a mass of 1 kg. By simply F = ma, Farreneheit 800 000 a= = = two m s”2 m some? 10 a few (b) Since the mass is fixed, the net push acting on it is zero. In order to flies horizontally, its velocity is 2 m s”2. 100 ( )? zero v? u (a) a = sama dengan 3. 6th = 5. 63 m s”2 capital t 6 The acceleration from the car is definitely 4. 63 m s”2. (c) (i) y-component of F1 = weight of mass = 10 D 9 y-component of F1 = F1 sin 30 F1 sin 30 = 10 D F1 = 20 N x-component of F1 sama dengan F1 cos 30 sama dengan 20 cos 30 = 17. 3 N (b) F = ma sama dengan 1500? 4. 63 = 6945 N The push provided by the automobile engine can be 6945 And. 10 (a) (ii) y-component of F2 = zero x-component of F2 sama dengan x-component of F1 = 17. 3 N