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4 years ago

A 8.1 kg object initially at rest is pushed down a 15.0 m tall hill. What is the speed of the object at the bottom of the hill?

Physics

1 answer:

topjm [15]4 years ago

8 0

Answer:

The velocity of the object at the bottom is, v = 17.15 m/s

Explanation:

Given data,

The initial velocity of the object, u = 0

The height of the hill, h = 15 m

Let 'S' be the distance of the slope of the hill and 'Ф' be the slope of the hill formed with the ground.

The acceleration due to gravity component along the slope is given by,

a = g Sin Ф

The distance of the slope since height 'h' of the hill is given,

s = h / Sin Ф

Using the III equation of motion,

v² = 2 as (∵ u = 0)

v² = 2 x g Sin Ф x h / Sin Ф

= 2 gh

Therefore,

Substituting the given values,

v = √(2x9.8x15)

= 17.15 m/s

Hence, the velocity of the object at the bottom is, v = 17.15 m/s

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yanalaym [24]

The fast lap is irrelevant to the question, because it didn't happen
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IF we can assume that Amy maintained a totally steady pace through
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3 years ago

a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(

horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.

Explanation:

We can find the velocity of the skier by energy conservation:

$E_{1} = E_{2}$

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

$mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}$ (1)

Where:

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h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

$v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}$

$v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s$

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!

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