Properties of matter are typically broken into two categories:physical and (fill in the blank)

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

$\terxt{time for compression}=\text{time for half revolution}$

$\terxt{time for compression}=\dfrac{1}{2}\times T$

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}$

Put the value into the formula

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60$

$\terxt{time for compression}=0.2\ sec$

We need to calculate the rate of internal energy

Using first law of thermodynamics

$U=Q-W$

$\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}$

Put the value into the formula

$\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)$

$\dfrac{\Delta U}{\Delta t}=6.8\ kW$

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

$\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}$

$\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}$

Put the value into the formula

$\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}$

$\Delta\theta=32.48^{\circ}$

Hence, The temperature change per compression stroke is 32.48°.

6 0

3 years ago

Friends who are your age and share the same interests are called

Pie

The answer is C, a peer group.

6 0

3 years ago

Read 2 more answers

Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c

Studentka2010 [4]

Answer:

30.96 m

Explanation:

If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:

d = v * Δt

v = 0.8*c = 0.8 * 3e8 = 2.4e8

Δt = ζ = 129 ns = 1.29e-7 s

d = 2.4e8 * 1.29e-7 = 30.96 m

This is the distance as measured by observer A.

3 0

3 years ago

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

where n=1,2,3 or ...n

L=Length of string

T=Tension

$\mu =$ Mass per unit length

When string is clamped at mid-point

Effecting length becomes $L'=0.5 L$

Thus new Frequency becomes

$f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}$

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz

4 0

3 years ago

Two loudspeakers are about 10 mm apart in the front of a large classroom. If either speaker plays a pure tone at a single freque

Yuliya22 [10]

Answer:

I hear points of low volume sound and points of high volume of sound.

Explanation:

This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.

Since their frequencies are similar, we should have beats of high and low frequency.

So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.

Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.

So, as you wander around the room, I should hear points of high and low sound across the room.

6 0

3 years ago