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strojnjashka [21]
3 years ago
7

Suppose the initial kinetic energy and final potential energy in an experiment are both zero. What can you conclude?

Physics
1 answer:
Kisachek [45]3 years ago
5 0

Initially, the experiment has only potential energy (since total energy is the sum of kinetic and potential energy). And at the end, the experment has only kinetic energy.

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Whats the first state discovered?​
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4 0
3 years ago
Please I need help! This is the last question I need for this assignment!
Ne4ueva [31]

Answer:

When the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

What is specific heat capacity?

Specific heat capacity is the quantity

of heat required to raise a unit mass of

a substance by 1 kelvin.

Specific heat capacity of water and sand

{<em>refer to the above attachment}</em>

Δθ = Q/mc

Thus, for an equal mass of water and sand, when the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

6 0
2 years ago
At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex
telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let F_{e} be the force of gravity exerted by the earth

and let F_{m} be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d,  is given by:

F = \frac{Gm_{1} m_{2} }{d^{2} }

Mass of the earth, m_{e} = 5.97 * 10^{24} kg

Mass of the moon, m_{m} = 7.348 * 10^{22} kg

Mass of the satellite, m_{s} = ?

F_{e}  = \frac{G*5.97 * 10^{24} M }{d^{2} }...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

F_{m}  = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }............................(2)

Equating equations (1) and (2)

\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }

(5.97 * 10^{24})(14.78 * 10^{16}  -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d +  5.97 * 10^{24}d^{2} =  7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\

Factorising out 10^{24}

1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0

Solving for d in the quadratic equation  above:

d = 346 * 10⁶ m

4 0
3 years ago
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