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Mekhanik [1.2K]
2 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]2 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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ankoles [38]
The human eye is not capable of seeing radiation wavelengths outside the visible Spectrum, i hope this answers your question :)
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Which formula correctly shows how to calculat the time taken, given the average velocity and the displacement
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We're so good here on Brainly, we can answer it
even WITHOUT seeing the choices.

     Time = (displacement) / (magnitude of average velocity) .
8 0
3 years ago
A ball is dropped from the top of the building.it initially moves at 4.0 m/s after 0.5 seconds it moves at 3.8m/s what force is
Ivahew [28]

The correct answer is

Air resistance

In fact, when a ball is in free fall, there are two forces acting on it:

- its weight (force of gravity), acting downward

- the air resistance, acting upward

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6 0
3 years ago
If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runn
yaroslaw [1]

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

<u>Slopes:</u>

Runner A:    y = 7.50x + 55

Runner B:    y = 7.90 x

sooooo

  7.50 x + 55 = 7.90 x

- 7.50 x          - 7.50 x

 55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 =  1086.25 m

3 0
2 years ago
An ice skater has a moment of inertia of 5.0 kg · m2when her arms are outstretched, and at this time she is spinning at 3.0 rev/
KatRina [158]

Answer:

Explanation:

Given

Initial Moment of Inertia I_1=5 kg-m^2

initial Spin N_1=3 rev/s

\omega _1=2\pi N_1=2\pi 3=6\pi rad/s

Final Moment Moment of Inertia I_2=2 kg-m^2

Conserving Angular momentum

L_1=L_2

I_1\omega _1=I_2\omega _2

5\times 6\pi=2\times \omega _2

\omega _2=15\pi rad/s

N_2=\frac{\omega _2}{2\pi }

N_2=\frac{15\pi }{2\pi}=7.5 rev/s

8 0
3 years ago
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