#3 priority but as many as you can do in the next hour please!!! . Andy and Bob
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Question
#3 priority but as many as you can do in the next hour please!!! . Andy and Bob are standing on a merry-go-round in such a way that the center of the merry-go- round, Andy, and Bob are aligned in a straight line, as shown below (A Andy; B-Bob; the dot is the center). The merry-go-round spins with constant angular velocity, completing 2 revolutions every 9.5 seconds. Bob is standing 3.7 meters away from the center of the merry-go-round, and the separation between Andy and Bob is 1.6 meters. What is the speed of Andy? 2. A planet with a mass of 8.50 × 10u3 kg moves in a (nearly-) circular orbit around a star with a mass of 2.50 x 10 kg. The separation between the planet and the star is measured as 2.10 x 10m. (a) What is the centripetal force acting on the planet? (b) How much time does it take for the planet to complete one revolution around the star? Give your answer in years. 3. A student places a 2.00-kg block on an inclined plane, as shown in Figure 1 (a), and measures the block's acceleration to be 1.40 m/s. The student then lowers the inclined plane to make it horizontal, and ties the block to a string that passes over a frictionless pulley and connects to a 1.00-kg hanging mass. This new arrangement is shown in Figure 1 (b). Calculate the expected acceleration of the block in the new arrangement. Note that the same block and plane are used in both arrangements (with the latter changing its inclination angle). 2.00kg 2.00 kg 0.0 1.00kg Figure 1 4. A 12.0-kg wooden crate rests on a horizontal loor. A man then pushes on the crate by exerting a horizontal force of magnitude 70.0 N. The crate does not move. The coefficient of static friction between the crate and the floor is 0.650. Three students argue about the situation, as shown below Select the student (s) whose statements are entirely correct. Student 1: "Since the crate is not moving, there must be a force of friction with magnitude 70.0 N pointing opposite to the force of the man." Student 2: "The force of static friction is equal to f, ,N. In this case, N = mg, so = (0.65)(12)(9.8)-76.4 N. Therefore, I don't agree that the force of friction is 70.0 N." Student 3: "I don't agree with either of you. We can't know what the force of static friction is; all that we can tell is that f, s 76.4N. A ball on the end of a string is swung at constant speed in a vertical circle of radius 60.0 cm. If its speed is at the top of its path, and (b) at the bottom of its path. 6. A 325-kg piano slides 4.00 m down an incline and is kept from accelerating by a man who pushes back on it parallel to the incline (the man is standing downhill from the piano, pushing uphill) The incline is tilted 29.0P above the horizontal, and the coetficient of kinetic friction between the piano and the incline is 0.400. Calculate the work done on the piano by (a) the man, (b) the friction force, (c) the force of gravity, and (d) the normal force. (e) What is the net work done on the piano?
Explanation / Answer
1) frequency of merry go round, f= 2/9.5 Hz = 0.2105 Hz
angular velocity,w =2*3.14*f = 1.321 rad/sec
distance between andy and centre, r= 3.7-1.6 = 2.1 m
andy's velocity, v= rw= 2.1*1.321= 2.7741 m/s
2) a)Centripetal force of the planet will be equal to the Gravitational force between them which is
F= GMm/R2 where, M is the mass of star, m is the mass of planet, R is the distance between them
F= 6.67*10-11 *2.5*1030*8.5*1023 / (2.1*1011) 2
F= 3.214*1021 N
b) mv2/R=3.214*1021 ; v=28178.84 m/s
Distance covered in one revolution= 2*3.14*R=1.318*1012 m
Time = distance/speed = 1.318*1012 /28178.84 = 46772684.75 seconds = 1.4831 years
3) Net force in first case(diagram a) is {angle of inclination is 20 degrees (180-(70+90))}
Fnet= mg*sin(70) - umgcos(70)
acceleration of the block can be given by, Fnet/m
a= g*sin(20) - ugcos(20) {where u is the coefficient of kinetic friction }
1.40 = 3.351 - u*9.208
u=0.211
The two equations in case two(diagram b) would be
T- u*2*g = 2*a
1*g - T = 1*a
where T is the tension in the wire and a is the acceleration of the system
Adding both the equations
g- 2ug = 3a
a= 1.888 m/s2
4) Student 1 is correct as friction force is self adjusting force whose value is equal to the applied force{this is true up until applied force is less than static friction force}
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