Study for Physics : exam 3A, 2A & homework

Q8 : Water flows over a section of Niagara Falls at a rate of 2.1 x 10^6 kg/s and falls 65 m. What is the power wasted by the waterfall?

p = 2.1 x 10^6 kg/s , h = 65 m, and t= 9.8 m/s ^2

Q6: A 500-kilogram sports car accelerates uniformly from rest, reaching a speed of 30 meters per second in 6 seconds. What distance has the car traveled during the 6 seconds?

s = s0 t + v0 t +12 a t2 =12 a t2 =12 vft t 2 =12 vf t=12 (30 m/s) (6 s) = 90 m .

Q6: A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 1.2m/s ^2. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the force of the elevator floor on the person?

0.789796

Q4: An escalator is 15.3 m long. If a person stands on the escalator, it takes 43.9 s to ride from the bottom to the top. if a person walks up the moving escalator with a speed of 0.545 m/s relative to the escalator, how long does it take the person to get to the top?

17.1233

Q3: A football is thrown upward at a (n) 49 angle to the horizontal. The acceleration of gravity is 9.8 m/s. To throw the ball a distance 72.9 m, what must be the initial speed of the ball.

26.859655380249

Q5: A car has as mass of 1.43x 10^3 kg. if the force acting on the car is 6.65 x 10^3 N to the east, what is the car's acceleration?

4.65035

Q11: With what velocity does it leave the ground?

As the ball rebounds, its motion is equivalent to dropping it from a height of h1 under the influence of gravity, and its speed will be v2 =p2 g h1 =q2(9.8 m/s2)(0.831 m) = 4.03579 m/s . The velocity is positive because of the upward direction.

Q7: You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You discover that the balloon bursts when you have pitched it to a height of 17 m. With what velocity did the balloon hit the ground? The acceleration of gravity is 9.8 m/s 2. The positive direction is up, so the velocity when it hits is negative.

Consider the second half of the motion with up positive:when the balloon was falling from rest (v0 =0 m/s) from the height h under the influence of gravity. The final speed is given by v2 f = v2 0 + 2 g h = 2 g h vf = ± p2 g h

Q2: Consider a toy car on an inclined ramp. The car is pushed up the ramp and released, it then coasts up the rap for a while, slows down to a complete stop, and eventually rolls back down the ramp. This problem focuses on the time period when the car is coasting up the ramp, after its released but before it stops.

In the absence of the friction force and the air drag, there are only two forces acting on the car after it is released. Both mg and N are constant, so the net force is also constant, in both magnitude and direction.

Q5: A racing car has a mass of 1970 kg. The acceleration of gravity is 9.8 m/s. What is its kinetic energy if it has a speed of 112 km/h?

K= ( 1/2 ) ^2 = (1/2)( 1970 kg ) = 9.53384 x 10^5

3A Q1 : Two marbles, one twice as massive as the other, are dropped from the same height. When they strike the ground, how does the kinetic energy of the more massive marble compare to that of the other marble?

KE + PE = constant. The marble with twice the mass has twice the gravitational PE at the same height,so it must have twice the KE when the PE goes to zero.

Q10: A tennis ball is dropped from 1.88 m above the ground. It rebounds to a height of 0.831 m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s2(Let down be negative.

Let : h = −1.88 m , h1 = 0.831 m , and g = −9.8 m/s2 Its final velocity is given by v1 = −p2 g h = −q2(−9.8 m/s2)(−1.88 m) = −6.07026 m/s . The velocity is negative because of the downward direction.

Q7: A mass of 1.8 kg lies on the friction-less table, pulled by another mass of 5.4 kg under the influence of Earth's gravity. What is the magnitude of the acceleration a of the two masses?

Let the direction of acceleration as indicated in the figure be positive. The net force on the system is simply the weight of m2. From Netwon's second law, Fnet = m2g = ( m1 + m2 )a.

2A Q1: A ball initially moves horizontally with velocity as shown. It is then struck by a stick. After leaving the stick, the ball moves vertically with a velocity, which has the same magnitude. Which vector best represents the direction of the average force tat the stick exerts on the ball?

The change in velocity is in the same direction as the acceleration during the same interval of time. And that's also, by Newton's 2nd law, the direction of the net average force during that time.

Q2: What is the position when t = 15 s?

The constant slope of the graph is 4 (9 m) 9 (5 s) . We want the position corresponding to thethird tic mark on the horizontal axis: 4(9 m) 9(5 s) = x 3(5 s) x = 3(5 s) 4(9 m) 9(5 s) = 12 m

Q3: After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far does the bird travel from the beginning (including the distance traveled to the first encounter)?

The distance remaining for the runner after the first encounter is L1 =23 L , and again the bird will fly 5 times as far as the runner until the next encounter. This pattern repeats over the entire original distance, so db = 5 L = 5 (7.5 km) = 37.5 km .

Q6 : Karen has a mass of 52.2 kg as she rides the up escalator at Woodley Park Station of the Washington D.C. Metro. Karen rode a distance of 16.7 m , one of the longest escalators in the world. The acceleration of gravity is 9.8 m/s. How much work did the escalator do on Karen if it has an inclination of 27.9?

The escalator raised Karen a height of h=dsin. So the work done on Karen by the escalator.

Q8: A (n) 1300 kg car is parked on a 4 incline. The acceleration of gravity is 9.8 m/s^2 . Find the force of friction keeping the car from sliding down the incline.

The force of friction equals the component of weight along the plane.

Q10 : A man cleaning his apartment pushes a vacuum cleaner with a force of magnitude 25.5 N. The force makes an angle of 18.4 with the horizontal floor. The vacuum is pushed 2.79 m to right along the floor.

The normal force, N, the weight of mg, and the upward component (25.5 N) sin 18.4 of the applied force do no work because they are perpendicular to the displacement.

HW Q1: What is the position represented by the fourth tic mark on the vertical axis?

The position is 4 times the scale factor: d = 4 (9 m) = 36 m .

Q3: Car B has the same mass as Car A but is going twice as fast. If the same constant force brings both cars to a stop, how far will car B travel in coming to rest as compared to the distance car A traveled?

The work done by friction in stopping the cars is W= f d = ( delta ) K. Since the force of friction is proportional to the mass of the car, as is K, then d depends on v. Thus, if v is doubled, then stopping distance increases by 4 times.

Q2 : Moe is carrying and supporting a large crate by exerting 1000 N. He is walking at constant velocity across a level, horizontal floor. If he covers a distance of 10 meters, how much work does he do on the crate?

When is Moe is carrying the crate as he moves at constant velocity, Moe only exerts a vertical force on the crate to counteract gravity. There is no horizontal force as the crate, moving at constant velocity, is not accelerating. Then, the force Moe exerts is perpendicular to the motion of the crate, and since this force has zero force component along the direction of motion, he is not doing any work on the crate when he moves at constant velocity.

Q12: If the tennis ball were in contact with the ground for 0.0122 s, find the acceleration given to the tennis ball by the ground.

While in contact with the ground, vo = v1 and vf = v2, so its acceleration while in contact with the ground is a =∆v∆t =vf − vo∆t =v2 − v1t =4.03579 m/s − (−6.07026 m/s)0.0122 s = 828.364 m/s2

Q7: On a frozen pond, a 5 kg sled is given a kick that imparts to it an initial speed of 2.3 m/s. The coefficient of kinetic friction between sled and ice is 0.16.

by the work of kinetic energy. = 1.68686m

Q9: An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees the car, the locomotive is 180 m from the crossing and its speed is 10 m/s. If the engineer's reaction time is 0.45 s, what should be the magnitude of the minimum deceleration to avoid an accident?

d = 180 m , v = vi = 10 m/s, and t = 0.45 s. While the engineer reacts, the train moves ∆d = v t = (10 m/s) (0.45 s) = 4.5 m forward, so it now has to decelerate to rest within a displacement of ∆x = d−∆d = 180 m−4.5 m = 175.5 m , and v2 f = v2 i + 2 a ∆x = 0 a =−v2i2 ∆x =−(10 m/s)2(175.5 m) = −0.2849 m/s2 which has a magnitude of 0.2849 m/s2

Q5: Ann is driving down a street at 61 km/h. Suddenly a child runs into the street. If it takes Ann 0.796 s to react and apply the brakes, how far will she have moved before she begins to slow down?

d = v t = (61 km/h) (0.796 s) 1 h 3600 s 1000 m 1 km = 13.4878 m .

Q4: A kilowatt-hour is a unit of

energy

Q8: During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.14 s, how high does it rise? The acceleration of gravity is 9.8 m/s2

g = 9.8 m/s2 y =12 g t2 =13 (9.8 m/s2)(3.07 s)2 = 46.182 m

Q4: A reconnaissance plane flies 511 km away from its base at 794 m/s, then flies back to its base at 1191 m/s. What is its average speed?

t1 = 511 km 794 m/s 1000 m km = 643.577 s and t2 = 511 km 1191 m/s 1000 m km = 429.051 s, so the total time is 1072.63 s. v¯ = total distance total time = 2 (511 km) 1072.63 s · 1000 m km = 952.8 m/s .

Q9 : A 3.38 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 14. N. The coefficient of kinetic friction is 0.113

to calculate the change in kinetic energy, ( delta ) k. The net force exerted on the block is the sum of the applied 14 N force and the frictional force is in the direction opposite displacement, it must be subtracted. The magnitude of the frictional force. Therefore the net force acting on the block is multiplying this constant force by the displacement, and using equation ( 1 ), we obtain since the initial velocity is zero.