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kaheart [24]
3 years ago
5

How is work related to potential energy and kinetic energy?

Physics
1 answer:
NemiM [27]3 years ago
3 0
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Featured snippet from the web
Work is the force on the object as it changes a distance. Interestingly, as work is done on an object, potential energy can be stored in that object. For example, if you carry a load up the stairs. Now that load will have potential energy that can be transformed into kinetic energy and so on
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Potential energy results from the ______or position of an object.
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Answer:

Mass

Explanation:

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A transport truck pulls on a trailer with a force of 600N [E]. The trailer pulls on the transport truck with a force
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These forces form a force pair. Use Newton's third law, and you see that the trailer pulls back at with the same force. The answer is d.
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A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold
tino4ka555 [31]

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



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3 years ago
Barnard’s Star is a red dwarf. It is located 5.9 light years from Earth. (One light year is the same as 9.46 trillion kilometers
mote1985 [20]
A star is located 5.9 light years from Earth.
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A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.
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Answer:

t = 0.192 \mu m

Explanation:

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here we know that

\mu = 1.79

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Shift = \frac{D(\mu - 1)t}{d}

Also we know that the fringe width of maximum intensity is given as

\delta x = \frac{\lambda D}{d}

now we have

\frac{D}{d} = \frac{\delta x}{\lambda}

now the shift is given as

Shift = \frac{(\mu - 1) t \delta x}{\lambda}

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Shift = 0.37 \delta x

here we have

0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}

now plug in all values in it

0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}

t = 0.192 \times 10^{-6} m

t = 0.192 \mu m

3 0
3 years ago
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