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gtnhenbr [62]
3 years ago
12

A sample of helium behaves as an ideal gas as energy is

Physics
1 answer:
alisha [4.7K]3 years ago
4 0

Answer:

0.0321 g

Explanation:

Let helium specific heat c_h = 5.193 J/g K

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

E_h = m_hc_h \Delta T = 20 J

where m_h is the mass of helium, which we are looking for:

m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g

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(a) 1200 rad/s

The angular acceleration of the rotor is given by:

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where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

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t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

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(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

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\omega_i = 2000 rad/s is the initial angular speed

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\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

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