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Mekhanik [1.2K]
3 years ago
15

A cylinder contains 250 L of hydrogen gas (H2) at 0.0^∘Cand a pressure of 10.0 atm. How much energy is required to raise the tem

perature of this gas to 25.0^∘C?

Physics
1 answer:
Over [174]3 years ago
8 0

Answer:

The amount of energy needed to raise the temperature of the cylinder by 25 °C is 23.3 KJ of heat.

Explanation:

The step by step calculation can be found in the attachment below. Thank you.

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"The total distance traveled divided by the time it takes to travel the distance"

That's actually a pretty good definition of average speed. <em>(A)</em>

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An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar
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A wave has a wave length of 20mm and a frequency of 5 hertz. What is the speed?
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Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha
ikadub [295]

We have that the spring constant is mathematically given as

k=2.37*10^{11}N/m

Generally, the equation for angular velocity is mathematically given by

\omega=\sqrt{k}{m}

Where

k=spring constant

And

\omega =\frac{2\pi}{T}

Therefore

\frac{2\pi}{T}=\sqrt{k}{n}

Hence giving spring constant k

k=m((\frac{2 \pi}{T})^2

Generally

Mass of earth m=5.97*10^{24}

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T=365 days

T=365*24*3600

Therefore

k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2

k=2.37*10^{11}N/m

In conclusion

The effective spring constant of this simple harmonic motion is

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8 0
3 years ago
A
True [87]

Answer:

\Delta T=3.615^{\circ}C is the drop in the water temperature.

Explanation:

Given:

  • mass of ice, m_i=14.7\ g=0.0147\ kg
  • mass of water, m_w=324\ g=0.324\ kg

Assuming the initial temperature of the ice to be 0° C.

<u>Apply the conservation of energy:</u>

  • Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

<u>Now from the heat equation:</u>

Q_i=Q_w

m_i.L=m_w.c_w.\Delta T ......................(1)

where:

L= latent heat of fusion of ice =333.55\ J.g^{-1}

c_w= specific heat of water =4.186\ J.g^{-1}.^{\circ}C^{-1}

\Delta T= change in temperature

Putting values in eq. (1):

14.7 \times 333.55=324\times 4.186\times \Delta T

\Delta T=3.615^{\circ}C is the drop in the water temperature.

8 0
3 years ago
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