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4 years ago

Which of the following is not a unit used to measure pressure?

Physics

2 answers:

Luden [163]4 years ago

7 0

Newtons are used to measure force

Ilya [14]4 years ago

4 0

Answer:

The answer is Newton.

Explanation:

In order to determine the unit that it is not used to measure pressure, we have to define the units for every option and then compare them.

Torr (symbol: Torr): is a unit of pressure based on an absolute scale, now defined as exactly 1/760 of a standard atmosphere (≈ 133.32 Pa).

Newton (symbol: N) is the International System of Units (SI) derived unit of force.

Atmosphere (symbol: atm): the standard atmosphere is a unit of pressure defined as 101325 Pa (1.013 bar)

Milimeter of mercury (symbol: mmHg): is a manometric unit of pressure, formerly defined as the extra pressure generated by a column of mercury one millimetre high, and currently defined as exactly 133.32 Pa.

Therefore, the physics unit that not represent a pressure measure is Newton.

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4 years ago

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3 years ago

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3 years ago

Consider a 10 gram sample of a liquid with specific heat 2 J/gK. By the addition of 400 J, the liquid increases its temperature

stich3 [128]

40 J/g is the heat of vaporization of the liquid.

Answer: Option D

<u>Explanation:</u>

Given that mass of liquid sample: m = 10 g

And, Specific heat of the liquid: S = 2 J/g K

Also, the increase in the temperature of the liquid, $\Delta T = T_{2}-T_{1} = 10 K$

Therefore, the total amount of heat energy required is given by:

$q_{1} = m \times S \times\left(T_{2}-T_{1}\right) = 10 \times 2 \times 10 = 200 J$

According to the given data in the question,

Total heat energy supplied, q = 400 J

Rest of heat would be $q_{2}=q-q_{1}=400-200=200 \mathrm{J}$

Now, 200 J vaporizes the mass, half of the liquid from full portion boiled away. So,

$m^{\prime} = \frac{10}{2} = 5 \mathrm{g}$

Latent heat of vaporization of the liquid is $L_{v}$ . It can be calculated as below,

$q_{2} = m^{\prime} L_{v}$

$L_{v} = \frac{q_{2}}{m^{\prime}} = \frac{200}{5} = 40 \mathrm{J} / \mathrm{g}$

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3 years ago