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Mademuasel [1]
3 years ago
13

B. Calculate the rock's gravitational potential energy at 50 m, 20 m, 1 m, and om high. Put the

Physics
1 answer:
klasskru [66]3 years ago
4 0

The gravitational potential energy of the rock is:

490 J at 50 m

196 J at 20 m

9.8 J at 1 m

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in the gravitational field.

For an object near the Earth's surface, the GPE is given by

GPE=mgh

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity at Earth's surface

h is the heigth of the object relative to the ground

For the rock in  this problem,

m = 1 kg

So its GPE at h = 50 m is

GPE(50)=(1)(9.8)(50)=490 J

The GPE at h = 20 m is

GPE(20)=(1)(9.8)(20)=196 J

While the GPE at h = 1 m is

GPE(1)=(1)(9.8)(1)=9.8 J

Learn more about gravitational potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
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Hi there!

We can use the work-energy theorem to solve.

Recall that:

\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

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Now, we can define work:

\large\boxed{W = Fdcos\theta}}

Now, plug in the values:

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Where is the safest spot in the house during a tornado, hail storm, and earthquake? Include separate answers.
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Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

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The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

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F=\frac{dp}{dt} ............................................(1)

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We know, momentum:

p=m.v

Now, equation (1) becomes:

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<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

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<em>& </em><u><em>Impulse:</em></u>

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That is not a question but not all scientific theories have stood the test of time
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