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3 years ago

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What is the gravitational potential energy of a 0.1 kg apple sitting on top of a 12.5 m flagpole? Note: GPE= mgh. m= mass, g= ac

celeration due to gravity (9.8 m/s^2),
h= objects height above the ground.

1 answer:

Pachacha [2.7K]3 years ago

6 0

<span>GPE= mgh = 0.1 kg * 9.8 m/s^2 * 12.5m = 12.25 J</span>

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Answer:

$a=2.8\ m/s^2$

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

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Answer:

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Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

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The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

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