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Cerrena [4.2K]
3 years ago
5

Newton's Law of Cooling says that the rate of cooling of an object is proportional to the difference between its own temperature

and the temperature of its surrounding. Write a differential equation that expresses Newton's Law of Cooling for this particular situation.
Physics
1 answer:
ASHA 777 [7]3 years ago
3 0

Answer:

\frac{dQ}{dt} =-hA\Delta T(t)

Explanation:Newton.s law of cooling states that the rate of cooling of an object is proportional to the difference between its own temperatures and temperature of its surroundings. Mathematically,

\frac{dQ}{dt} =-hA [T(t)-T(s)]\\

\frac{dQ}{dt} =-hA\Delta T(t)

where Q is the heat transfer

h is heat transfer coefficient

A is the heat transfer surface area

T is the temperature of the object's surface

T(s) is the temperature of the surroundings

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If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th
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Answer:

\frac{K.E_r}{K.E}=2.875

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K.E_r=(\gamma-1)mc^2 ...........(1)

where,

\gamma = relativistic factor, given as; \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

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

Dividing the equation (1) by (2) we get

\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}

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substituting the values in the equation we get,

\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}

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\frac{K.E_r}{K.E}=2.875

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Explanation:

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