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求救 3條phy 題目
發問:
Question 1a. A force of 33 N is applied to a 4 kg toy car, it moves along a plane. If the frictional forceof the plane is 5 N, find the acceleration of the toy car.b. The mass of a toy car is 2 kg, it is pulled by a 7 N force.If the acceleration of the car is 1.5 ms-2, find the frictional force.c. A... 顯示更多 Question 1 a. A force of 33 N is applied to a 4 kg toy car, it moves along a plane. If the frictional force of the plane is 5 N, find the acceleration of the toy car. b. The mass of a toy car is 2 kg, it is pulled by a 7 N force. If the acceleration of the car is 1.5 ms-2, find the frictional force. c. A boy throws a ball to a maximum height 5 m above the ground vertically. Calculate the initial velocity of the ball and time taken to reach the maximum height. d. A box of 5 kg slides down on the inclined plane. (i) Find the downward force along the plane if no friction. (ii) If friction is 3N, find the acceleration along the plane. Question 2 a. A bus moves from rest and takes 10 sec. to cover a distance of 300 m. Calculate the acceleration of the bus. b. A ball is thrown up vertically with an initial velocity of 5 ms-1. Calculate the maximum height it reaches and the time taken for it to reach that height. c. A driver brakes a bus which is traveling with an initial velocity of 20 ms-1 . It decelerates at 6 ms-2. Calculate the time taken and distance moved when it comes to stop. d. A car has a mass of 5000 kg. It accelerates from rest to a final velocity of 5 ms-1, during which it travels a distance of 40 m. If the total friction acting on the car is 3000 N. (i) Find the total force developed by the car’s engine; (ii) Also find the energy output from the car’s engine. Question 3 (a) A 4 kg box is pulled by a 10 N force. If the friction is 2 N, find the acceleration b) A 3 kg ball at 6 m height falls down to the ground, (i) Find the initial P.E. at 6 m height. (ii) If no energy loss, find the velocity when it just hits the ground. (iii) If the energy loss in air resistance is 15 J, find the velocity when it just hit the ground. (c) A box of 6 kg is released from rest as shown in Fig. Q3c. It slides down the inclined plane with a friction of 10 N. Find the time for the box to reach the ground level. 更新: (d) Box A of 60 kg sits on 3 m from the pivot, and box B of 40 kg sits on other side of the lever. If the lever is at balanced condition, find the distance, d.
最佳解答:
1. (a) acceleration = (33-5)/4 m/s2 (b) Friction = (7 - 2x1.5) N (c) Initial speed, u = square-root[2 x 10 x5] m/s Time taken = 2 x5/u = 10/square-root[2 x 10 x5] s (d) (i) This depends on the angle of inclination of the plane downward force = 50.sin(a) N, where a is the angle of inclination (ii) acceleration = (50sin(a) -3)/5 m/s2 2. (a) acceleration = (2 x 300/10x10) m/s2 (b) Height reached = 5x5/2x10 m Time taken = 5/10 s (c) There is not enough data to calculate. (d)(i) Force ={ [(1/2)(5000)(5x5)/40] + 3000} N (ii) Energy output = [(1/2)(5000)(5x5) + 3000x40] J 3. (a) Acceleration = (10-2)/4 m/s2 (b) (i) PE = 3x10x6 J (ii) velocity = square-root[2x10x6] m/s (iii) velocity = square-root[(2/3)x(3x10x6-15)] m/s (c) No figure is given. (d) d = 60x3/40 m Hope you could exercise some effort to think why the calculations are so.
其他解答:
1. abcd Fnet=ma 33-5=4(a) 28=4a a=7 m/s^2 Fnet=ma 7-f = 2(1.5) 7-f = 3 f = 4N s=5 g=-10 U=? v^2-u^2=2(a)(s) 0^2-u^2=2(10)(5) u^2=100 u=10 m/s apply s=ut+1/2(a)(t^2) s=ut+1/2(-10)(t^2) 5=(10)t+(-5)(t^2) 2010-03-24 17:35:44 補充： d I) Fnet=ma = 5(-10) =-50N 2a u=0 t=10 s=300 s=ut+1/2(a)(t^2) 300=(0)(10)+1/2(a)(100) 300=50a a = 6m/s^2 2b u=5 a=10 v^2-u^2=2(a)(s) 0-25 = 20s s= -1.25m s=ut+1/2(a)(t^2) -1.25=5t+1/2(10)(t^2) -1.25=5t+5t^2 -1.25=5t(1+t) -0.25=2t t=-0.125s 3a Fnet=ma 10-2=4a 8=4a a=2m/s^2
求救 3條phy 題目
發問:
Question 1a. A force of 33 N is applied to a 4 kg toy car, it moves along a plane. If the frictional forceof the plane is 5 N, find the acceleration of the toy car.b. The mass of a toy car is 2 kg, it is pulled by a 7 N force.If the acceleration of the car is 1.5 ms-2, find the frictional force.c. A... 顯示更多 Question 1 a. A force of 33 N is applied to a 4 kg toy car, it moves along a plane. If the frictional force of the plane is 5 N, find the acceleration of the toy car. b. The mass of a toy car is 2 kg, it is pulled by a 7 N force. If the acceleration of the car is 1.5 ms-2, find the frictional force. c. A boy throws a ball to a maximum height 5 m above the ground vertically. Calculate the initial velocity of the ball and time taken to reach the maximum height. d. A box of 5 kg slides down on the inclined plane. (i) Find the downward force along the plane if no friction. (ii) If friction is 3N, find the acceleration along the plane. Question 2 a. A bus moves from rest and takes 10 sec. to cover a distance of 300 m. Calculate the acceleration of the bus. b. A ball is thrown up vertically with an initial velocity of 5 ms-1. Calculate the maximum height it reaches and the time taken for it to reach that height. c. A driver brakes a bus which is traveling with an initial velocity of 20 ms-1 . It decelerates at 6 ms-2. Calculate the time taken and distance moved when it comes to stop. d. A car has a mass of 5000 kg. It accelerates from rest to a final velocity of 5 ms-1, during which it travels a distance of 40 m. If the total friction acting on the car is 3000 N. (i) Find the total force developed by the car’s engine; (ii) Also find the energy output from the car’s engine. Question 3 (a) A 4 kg box is pulled by a 10 N force. If the friction is 2 N, find the acceleration b) A 3 kg ball at 6 m height falls down to the ground, (i) Find the initial P.E. at 6 m height. (ii) If no energy loss, find the velocity when it just hits the ground. (iii) If the energy loss in air resistance is 15 J, find the velocity when it just hit the ground. (c) A box of 6 kg is released from rest as shown in Fig. Q3c. It slides down the inclined plane with a friction of 10 N. Find the time for the box to reach the ground level. 更新: (d) Box A of 60 kg sits on 3 m from the pivot, and box B of 40 kg sits on other side of the lever. If the lever is at balanced condition, find the distance, d.
最佳解答:
1. (a) acceleration = (33-5)/4 m/s2 (b) Friction = (7 - 2x1.5) N (c) Initial speed, u = square-root[2 x 10 x5] m/s Time taken = 2 x5/u = 10/square-root[2 x 10 x5] s (d) (i) This depends on the angle of inclination of the plane downward force = 50.sin(a) N, where a is the angle of inclination (ii) acceleration = (50sin(a) -3)/5 m/s2 2. (a) acceleration = (2 x 300/10x10) m/s2 (b) Height reached = 5x5/2x10 m Time taken = 5/10 s (c) There is not enough data to calculate. (d)(i) Force ={ [(1/2)(5000)(5x5)/40] + 3000} N (ii) Energy output = [(1/2)(5000)(5x5) + 3000x40] J 3. (a) Acceleration = (10-2)/4 m/s2 (b) (i) PE = 3x10x6 J (ii) velocity = square-root[2x10x6] m/s (iii) velocity = square-root[(2/3)x(3x10x6-15)] m/s (c) No figure is given. (d) d = 60x3/40 m Hope you could exercise some effort to think why the calculations are so.
其他解答:
1. abcd Fnet=ma 33-5=4(a) 28=4a a=7 m/s^2 Fnet=ma 7-f = 2(1.5) 7-f = 3 f = 4N s=5 g=-10 U=? v^2-u^2=2(a)(s) 0^2-u^2=2(10)(5) u^2=100 u=10 m/s apply s=ut+1/2(a)(t^2) s=ut+1/2(-10)(t^2) 5=(10)t+(-5)(t^2) 2010-03-24 17:35:44 補充： d I) Fnet=ma = 5(-10) =-50N 2a u=0 t=10 s=300 s=ut+1/2(a)(t^2) 300=(0)(10)+1/2(a)(100) 300=50a a = 6m/s^2 2b u=5 a=10 v^2-u^2=2(a)(s) 0-25 = 20s s= -1.25m s=ut+1/2(a)(t^2) -1.25=5t+1/2(10)(t^2) -1.25=5t+5t^2 -1.25=5t(1+t) -0.25=2t t=-0.125s 3a Fnet=ma 10-2=4a 8=4a a=2m/s^2
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