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4 years ago

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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

uation PV = C, where C is a constant. Suppose that at a certain instant the volume is 400 cm3, the pressure is 160 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?

1 answer:

Zigmanuir [339]4 years ago

3 0

Answer:

100 cm^3/min

Explanation:

PV = C

V = 400 cm^3

P = 160 kPa

dP/dt = + 40 kPa/min

According to the question

PV = C

Differentiate both sides with respect to time

P x dV/dt + V x dP/dt = 0

160 x dV/dt = - 400 x 40

dV/dt = - 100 cm^3/min

The volume decreases by the rate of 100 cm^3/min.

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bogdanovich [222]

Answer:

When they are approaching each other

$f_a = 2228.7 \ Hz$

When they are passing each other

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When they are retreating from each other

$f_a = 1980.7 Hz$

Explanation:

From the question we are told that

The velocity of car one is $v_1 = 13.0 m/s$

The velocity of car two is $v_2 = 7.22 m/s$

The frequency of sound from car one is $f_e = 2.10 kHz$

Generally the speed of sound at normal temperature is $v = 343 m/s$

Now as the cars move relative to each other doppler effect is created and this can be represented mathematically as

$f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]$

Where $v_s$ is the velocity of the source of sound

$v_o$ is the velocity of the observer of the sound

$f_o$ is the actual frequence

$f_a$ is the apparent frequency

Considering the case when they are approaching each other

$f_a = f_o [\frac{v + v_o}{v - v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ]$

$f_a = 2228.7 \ Hz$

Considering the case when they are passing each other

At that instant

$v_o = v_s = 0m/s$

$f_o = f_e$

$f_a = f_o [\frac{v }{v } ]$

$f_a = f_o$

Substituting value

$f_a = 2100Hz$

Considering the case when they are retreating from each other

$f_a = f_o [\frac{v - v_o}{v + v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 - 7.22}{343 + 13} ]$

$f_a = 1980.7 Hz$

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