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3 years ago

Which of the following objects have gravitational potential energy?

Physics

2 answers:

Jlenok [28]3 years ago

6 0

Answer:

an apple on a tree branch 3m above the ground

Explanation:

Brut [27]3 years ago

3 0

Answer:

The apple

Explanation:

The apple has gravitational potential energy because it is just sitting there but nothing is pushing it up, it is not sitting on something. This means that it has a lot of gravitational potential energy as nothing is pushing it up.

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A ball is thrown forward at 5 m/s. How would the path of the ball differ on Earth than on the moon? The ball would follow a curv

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Answer:

Explanation:

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3 years ago

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Una tortuga se desplaza en línea recta 100cm al norte y luego 80cm al este, determine su desplazamiento

valentinak56 [21]

Answer:

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5 0

2 years ago

Write the terms involved in s=ut+1\2

cricket20 [7]

s = displacement; u = initial velocity; t = time of motion

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2 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

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

$A=\sqrt{l^2-b^2}b$

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

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

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

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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2 years ago

A paper pinwheel is spinning in the wind. Which statement is correct about the forces responsible for the rotation?

VikaD [51]

Answer:

Only the perpendicular component of gravity is responsible for the rotation because wind points toward the pivot.

Explanation:

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3 years ago