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3 years ago

Does a falling rock have potential or kinetic energy

2 answers:

5 0

A falling rock has both potential and kinetic energy. But since the rock is losing height and potential energy is proportional to height, the rock is losing potential energy and gaining kinetic energy due to the increasing speed downwards.

Neko [114]3 years ago

3 0

depends t what stage in the fall it is. If it is at the peak, it is fully potential. If it is in the middle, it has both. If it is at the bottom of the fall, it is completely kinetic

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Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th

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Explanation:

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Which of the following is a possible use for nuclear energy?

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3 years ago

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LUCKY_DIMON [66]

Answer:

1. 37.8J

2. 18 Billion Joules, 18 Gigajoules

3. 9.81 Billion Joules, 9.81 Gigajoules

Explanation:

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A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

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3 years ago