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worty [1.4K]
3 years ago
10

gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure

Physics
1 answer:
Mamont248 [21]3 years ago
7 0

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be '\text{P}_2'.

As we know that Boyle's law formula states that;

                    P_1 \times V_1 = P_2 \times V_2

where, P_1 = original pressure of gas in the container = 310 mm hg

           P_2 = required new pressure

            V_1 = volume of gas in the container = 185 ml

            V_2 = desired new volume of the gas = 74 ml

So,  P_2 = \frac{P_1 \times V_1}{V_2}  

       P_2 = \frac{310 \times 185}{74}

            =  775 mm hg

Hence, the required new pressure is 775 mm hg.

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The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh
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The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, P_{headlight} = 38 W

The power at which the kick starter operates in parallel, P_{kick \ starter} = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/P_{headlight}  = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/P_{kick \ starter} = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

P_{headlight} = I² × R₁

∴ P_{headlight} = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, P_{headlight, S} ≈ 0.134 W

P_{kick \ starter} = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, P_{kick \ starter, S} ≈ 2.123 W

The total power they will consume, P_{Total} = P_{headlight, S} + P_{kick \ starter, S}

Therefore;

P_{Total} ≈ 0.134 W + 2.123 W = 2.257 W

4 0
3 years ago
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