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gogolik [260]
3 years ago
13

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C

where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Physics
2 answers:
DiKsa [7]3 years ago
5 0

Answer:

\frac{dV}{dt}=21.21cm^3/min

Explanation:

We are given that

PV^{1.4}=C

Where C=Constant

\frac{dP}{dt}=-7KPa/minute

V=420 cubic cm and P=99KPa

We have to find the rate at which the  volume increasing at this instant.

Differentiate w.r.t t

V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0

Substitute the values

(420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0

1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)

\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}

\frac{dV}{dt}=21.21cm^3/min

kupik [55]3 years ago
3 0

Answer:

\dot V=2786.52~cm^3/min

Explanation:

Given:

initial pressure during adiabatic expansion of air, P_1=99~kPa

initial volume during the process, V_1=420~cm^3

The adiabatic process is governed by the relation PV^{1.4}=C ; where C is a constant.

Rate of decrease in pressure, \dot P=7~kPa/min

Then the rate of change in volume, \dot V can be determined as:

P_1.V_1^{1.4}=\dot P.\dot V^{1.4}

99\times 420^{1.4}=7\times V^{1.4}

\dot V=2786.52~cm^3/min

\because P\propto\frac{1}{V}

\therefore The rate of change in volume will be increasing.

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\theta=29.84^{0}

For a larger angle for the first polarizer:

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I_{2} =I_{1} cos^{2}\theta

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A 40 cm wire with a radius of 3 cm is oriented along the y axis and carries a current of 2 A. What is the magnitude of the magne
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Substitute the value,

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