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3 years ago

PV = nRT, is the a. equation of state of an ideal gas

Physics

1 answer:

romanna [79]3 years ago

3 0

Ideal gas law:

PV = nRT

P = pressure, V = volume, n = # of moles, R = gas constant, T = temperature

Equipartition theorem:

Each degree of freedom that a molecule has adds 0.5kT to its total internal energy where k = Boltzmann's constant and T = temperature

2nd law of thermodynamics:

A set of governing principles that restrict the direction of net heat flow (always hot to cold, heat engines are never 100% efficient, entropy always tends to increase, etc)

Clearly the answer is Choice A

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Please help with this

sertanlavr [38]

Answer:I have to say 56

Explanation: because it is going up by 8

6 0

3 years ago

How much heat is needed to raise the temperature of 2 kg of copper from 20º to 30ºC. The specific heat of copper is 390 J/kgºC.

ryzh [129]

Answer:

7800 J

Explanation:

Heat needed = mass of copper x specific heat of copper x change in temperature

Change in temperature = 30ºC - 20ºC = 10ºC

Specific heat of copper = 390 J/kgºC

Mass of copper = 2 Kg

Substituting the given values in above equation, we get –

Heat needed = 2 Kg x 390 J/kgºC x 10ºC

= 7800 J

3 0

4 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

What is the speed of a transverse wave in a rope of length 2 meters and mass 0.06

IgorLugansk [536]

Answer:

16.3 wave speed.

Explanation:

8 0

3 years ago

Please help, question is timed...

Katen [24]

Answer:

Explanation:

50+3*5

8 0

3 years ago