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DanielleElmas [232]
3 years ago
8

A ball with a mass of 275 g is dropped from rest, hits the floor, and re-bounds upward. If the ball hits the floor with a speed

of 3.28 m/s, re-bounds with a speed of 1.63 m/s, and is in contact with the floor for 2.80 ms, determine the following.
A. Magnitude of the change in the ball's momentum.
B. Change in the magnitude of the ball's momentum.
C. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum.
D. The magnitude of the change in the ball's momentum
Physics
1 answer:
disa [49]3 years ago
3 0

Answer:

a) \Delta p = 1.350\,\frac{kg\cdot m}{s}, b) \Delta p' = -0.454\,\frac{kg\cdot m}{s}, c) D. The magnitud of the change in the ball's momentum.

Explanation:

a) The magnitude of the change in the ball's momentum is:

\Delta p = (0.275\,kg)\cdot \left[\left(1.63\,\frac{m}{s} \right)-\left(-3.28\,\frac{m}{s} \right)\right]

\Delta p = 1.350\,\frac{kg\cdot m}{s}

b) The change in the magnitude of the ball's momentum:

\Delta p' = (0.275\,kg)\cdot \left[(1.63\,\frac{m}{s} )-(3.28\,\frac{m}{s} ) \right]

\Delta p' = -0.454\,\frac{kg\cdot m}{s}

c) The magnitude of the change in the ball's momentum is more directly related to the net force acting on the ball, as it measures the effect of the force on change in ball's motion at measured time according to the Impact Theorem. So, the right answer is option D.

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Diffrence between:- movable pulley and fixed pully​
kotykmax [81]

Answer:

fixed pulley: A pulley system in which the pulley is attached to a fixed point and the rope is attached to the object. ... movable pulley: A pulley system in which the pulley is attached to the object; one end of the rope is attached to a fixed point and the other end of the rope is free.

Explanation:

8 0
2 years ago
Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe
uranmaximum [27]

Answer:

<em>a) 105935.7 Pa</em>

<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = \pi r^2h

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x 0.13^2 x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

7 0
3 years ago
How long would it take to go 11 miles at 22 mph?
harkovskaia [24]
It would tack about 3.2 h

5 0
3 years ago
If an organism can not reproduce,is it still considered living?why?
zlopas [31]
The ability to reproduce is essential for a species, but not for an individual. There are some people that can't reproduce, but they're still alive!
6 0
3 years ago
Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter
valentinak56 [21]

Wee can use here kinematics

as we know that

y = v*t + \frac{1}{2} at^2

for shorter tree we know that

y = 0 + \frac{1}{2}*9.8 * 2^2

y = 19.6 meter

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

y = v*t + \frac{1}{2} at^2

39.2 = 0 + \frac{1}{2}*9.8 * t^2

t = 2.83 s

so the time taken is 2.83 s

4 0
3 years ago
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