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drek231 [11]
2 years ago
12

A piston-cylinder chamber contains 0.1 m3 of 10 kg R-134a in a saturated liquid-vapor mixture state at 10 °C. It is heated at co

nstant temperature until the piston expands and the final volume of R-134a is 1 m3 . What is (a) the final pressure and (b) the total energy transferred to R-134a for this process?
Physics
1 answer:
vaieri [72.5K]2 years ago
6 0

Answer:

(A) 10132.5Pa

(B)531kJ of energy

Explanation:

This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.

Given

m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³

P1 = 101325Pa. M = 102.03g/mol

P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa

(B) Energy is transfered by the r134a in the form of thw work done in in expansion

W = nRTIn(V2/V1)

n = m / M = 10000/102.03 = 98.01mols

W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)

= 531kJ.

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Answer:

any amount of force will do it as time is not mentioned here

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the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be
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D:the electrons from being attracted to the grid instead of the anode
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1 year ago
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Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
1 year ago
A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.
kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

F_f = \mu N = \mu mg

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s

3 0
1 year ago
When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
Degger [83]

Answer:

(C) 4 beats per second.

Explanation:

As we know that the no of beats can be calculated as.

No. of beats is equal to difference in the tuning forks frequencies.

So,

n= \nu _{1}- \nu _{2}.

Substitute the values of frequencies of 2 tuning forks in the above equation.

n=(260 Hz-256 Hz)\\n=4

Therefore the number of beats per second will be hear by the observer is 4 beats per second.

3 0
2 years ago
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