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vovangra [49]
3 years ago
7

Identify what happens to the kinetic energy (KE) and gravitational potential energy (PE) of a rock as it falls.

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

KE increases and PE decreases.

Explanation:

<u>POTENTIAL ENERGY (PE)</u>:

The potential energy of the rock depends upon the height of the rock, as given by the following formula:

P.E = mgh

Therefore, the potential energy will be highest at the initial point due to the highest height attained by rock. As the rock falls, its height will decrease as a result its <u>potential energy will also decrease</u>.

<u>KINETIC ENERGY (KE)</u>:

The kinetic energy of the rock depends upon the velocity of the rock, as given by the following formula:

K.E = \frac{1}{2}mv^2

Therefore, the kinetic energy will be zero at the initial point due to the zero velocity of the rock. As the rock falls, its velocity will increase as a result its <u>kinetic energy will also increase</u>.

Hence, the correct option will be:

<u>KE increases and PE decreases.</u>

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Explanation:

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3 years ago
The motor on a helicopter turns at an angular speed of 6.2 x 102 revolutions per minute. (a) Express this angular speed in radia
arsen [322]

Answer:

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Explanation:

(a)

  \dfrac{6.2\cdot 10^2\,\text{rev}}{60\,\text{s}}\times\dfrac{2\pi\,\text{rad}}{\text{rev}}=\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\approx\boxed{64.93\,\text{rad/s}}

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(b)

  d=r\theta=(3.0\,\text{m})(2.0\cdot 10^2\,\text{s})\left(\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\right)=124\pi\cdot 10^2\,\text{m}=\boxed{38\,956\,\text{m}}

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3 years ago
Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an
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Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

\frac{1}{f} = \frac{1}{265} + \frac{27}{10}

  = \frac{10+7155}{2650}

\frac{1}{f}  = \frac{7165}{2650}

⇒ f = \frac{2650}{7165}

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The focal length of the eye is 0.37 cm.

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When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p
const2013 [10]

Answer: Fourth option. It increased by a factor of 3.

Solution:

m1=1.0 kg

Cylinder's gravitational potential energy: Ep=m*g*h

Ep1=(1.0 kg)*g*h

Ep1=g*h

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Replacing g*h by Ep1 in the equation above:

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4 years ago
Read 2 more answers
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