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Greeley [361]
2 years ago
10

You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball

descend together a distance of 0.150 m.
How much work do your hands do on the volleyball in the process of stopping it?
What is the magnitude of the force (assumed constant) that your hands exert on the volleyball?
Physics
1 answer:
sesenic [268]2 years ago
3 0

Explanation:

The work done equals the change in energy.

W = ΔKE

W = 0 − ½mv²

W = -½ (0.270 kg) (-7.50 m/s)²

W = -7.59 J

Work is force times displacement.

W = Fd

-7.59 J = F (-0.150 m)

F = 50.6 N

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garri49 [273]
We can use kinematics here if we assume a constant acceleration (not realistic, but they want a single value answer, so it's implied). We know final velocity, vf, is 1.0 m/s, and we cover a distance, d, of 0.47mm or 0.00047 m (1m = 1000mm for conversion). We also can assume that the flea's initial velocity, vi, is 0 at the beginning of its jump. Using the equation vf^2 = vi^2 + 2ad, we can solve for our acceleration, a. Like so: a = (vf^2 - vi^2)/2d = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, not bad for a flea!
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A moving sidewalk has a velocity of 1.7m/s north. if a man walks 1.1m/s, how long does it take him to travel 15m north in relati
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A zebra runs across a field at a constant speed of 14m/s how far does the zebra go in 8 seconds?
Ratling [72]

Answer:

112m/s

Explanation:

14x8=112 therefore meaning the zebra would run 112m/s

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A policeman kicks in a door with a force of 4500 N. What force does the door apply to the policeman’s leg?
Soloha48 [4]

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2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

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\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

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K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
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