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3 years ago

The boiling point of a substance is _72 degree Celsius. This temperature will be equivalent to Kelvin scale is-------.

Physics

1 answer:

Vlada [557]3 years ago

3 0

Answer:

345 K

Explanation:

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object.

Generally, it is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).

Given the following data;

Boiling point = 72°C

To convert the temperature in degree Celsius to Kelvin, we would use the following mathematical expression;

Kelvin = 273 + °C

Substituting into the formula, we have;

Kelvin = 273 + 72

Kelvin = 345 K

Therefore, the temperature of 72°C will be equivalent to 345 K on the Kelvin scale.

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Answer:

No the given statement is not necessarily true.

Explanation:

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For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

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Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

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3 years ago

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Answer:

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= 7.4 m

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v = √2ugd

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3 years ago