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Alecsey [184]
3 years ago
11

Two cylinders A and B at the same temperature contain the same quantity of the same kind of gas. Cylinder A has three times the

volume of cylinder B. What can you conclude about the pressures the gases exert? Show all workings. [15 marks] a. b. c. d. The density of gas in A and B are the same. The pressure in A is three times the pressure in B. The pressures in A and B must be equal. The pressure in A must be one-third the pressure in B.
Physics
1 answer:
astraxan [27]3 years ago
3 0

Answer:

so pressure in A must be one third the pressure in B

Explanation:

We shall apply gas law to the cylinders A and B . Since their quantity are same so their no of  mole will also be same .

For cylinder A

Temperature T , volume 3V , pressure  P₁ , no of mole = n

so

P₁ X 3V = n R T

For cylinder B

Temperature T , volume V , pressure  P₂ ,no of  mole = n

so

P₂ X V = n R T

From the two equation above

P₁ X 3V = P₂ X V

\frac{P_1}{P_2}=\frac{1}{3}

P₁ = P₂ / 3

so pressure in A must be one third the pressure in B

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Answer:

  0.017 N

Explanation:

The relevant relation is ...

  F = GMm/r²

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Filling in the given numbers, we find the force to be ...

  F = (6.67408 × 10^-11 m^3·kg^-1·s^-2)(8.7 × 10^20 kg)(77 kg)/(1.6 × 10^7 m)^2

where m in this expression is the unit "meters".

  F = 6.67408 · 8.7 · 77/2.56 × 10^(-11 +20 -2·7) N ≈ 0.017 N

The asteroid exerts a force of about 0.017 N on Sally.

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<em>Additional comment</em>

That's about 0.000023 times the force of Earth's gravity.

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<u>Explanation:</u>

<u>Given:</u>

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You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,

f^{\prime}=0.959 f

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If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,

                     f^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) \times f

Where,

f^' - Observed frequency

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v – Velocity of sound waves

v_0 – Velocity of observer

v_s - velocity of source

When source moves away from an observer at rest (v_{0} = 0), the equation would be

                        f^{\prime}=\left(\frac{v}{v-\left(-v_{s}\right)}\right) \times f

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By substituting the known values, we get

            0.959=\left(\frac{343}{343+v_{s}}\right)

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           0.959(343)+0.959\left(v_{s}\right)=343

           328.937+0.959 v_{s}=343

           0.959 v_{s}=343-328.937=14.063

           v_{s}=\frac{14.063}{0.959}=14.66 \mathrm{m} / \mathrm{s}

Approximately 15 m/s is the speed of the car.

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