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vladimir2022 [97]
3 years ago
13

How much energy (in joules) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature?

Physics
2 answers:
viva [34]3 years ago
8 0

The energy released when a certain amount of a substance condenses to a liquid is given by:

Q=m \lambda_v

where m is the mass of the substance and \lambda_v is the latent heat of vaporization.

In this problem, the mass is m=0.06 kg, while the latent heat of vaporization of mercury is \lambda_v=296 kJ/ kg. Therefore, if we apply the formula we get:

Q=(0.06 kg)(296 kJ/ kg)=17.76 kJ=17760 J

So, the correct answer is

C. 17,705.1 J

zvonat [6]3 years ago
7 0
The answer is ....  C. 17,705.1 J
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Answer:

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Explanation:

given,

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magnitude of momentum after collision = ?

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final momentum after collision

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P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

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3 years ago
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8_murik_8 [283]

Answer:

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Explanation:

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E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J

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\frac{4ev}{1.549802445eV}=2.508974118

Rounding to the next integrer: 3.

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