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Artist 52 [7]
2 years ago
13

When a quantity of monatomic ideal gas expands at a constant pressure of 4.00×104pa, the volume of the gas increases from 2.00×1

0−3m3 to 8.00×10−3m3?
Physics
2 answers:
3241004551 [841]2 years ago
7 0
This can be verified if we know the values of the initial (T1) and final (T2) temperatures. We use the ideal gas equation for this: PV=RT.

P1V1=RT1
(40000 Pa)*(0.002 m^3) = (8.314 m3Pa/molK)(T1)
T1 = 9.62 K


P2V2=RT2
(40000 Pa)*(0.008 m^3) = (8.314 m3Pa/molK)(T2)
T2 = 33.5 K

Thus, this is true if the monoatomic ideal gas is heated from 9.62 K to 33.5 K at constant pressure.
Semmy [17]2 years ago
5 0
Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^<span> 4) = </span>8.00×10^−3. That is how you get your answer
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Explanation:

Given: Mass = 250 kg

Initial temperature = 16^{o}C

Final temperature = 38^{o}C

Specific heat capacity = 4200 J/kg^{o}C

Formula used to calculate the energy is as follows.

q = m \times C \times (T_{2} - T_{1})

where,

q = heat energy

m = mass of substance

C = specific heat capacity

T_{1} = initial temperature

T_{2} = final temperature

Substitute the values into above formula as follows.

q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C

As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.

\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J

Thus, we can conclude that the energy incident on the solar panel during that day is 9.24 \times 10^{7} J.

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2 years ago
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a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin
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Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

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

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

\Delta t is the time interval

For the fan in this problem,

\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s

Substituting,

\alpha = \frac{180-0}{4}=45 rpm/s

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

\omega' = \omega_i + \alpha t

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Substituting,

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2 years ago
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