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SVEN [57.7K]
3 years ago
10

​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th

e pressure is 1.0 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air.
Determine the work and heat transfer, in kJ.

For an ideal gas, the specific volume is equal to:

A: the gas constant R times the gage temperature divided by the gage pressure.

B: the universal gas constant times the absolute temperature divided by the gage pressure.

C: the gas constant R times the absolute temperature divided by the absolute pressure.

D: the universal gas constant times the absolute temperature divided by the absolute pressure.
Physics
1 answer:
vlada-n [284]3 years ago
5 0

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

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Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, v_y, of the package just before landing is given by the relation;

v_y² = u² + 2·g·h

u = 0 m/s

∴ v_y² = 0 + 2×9.81×1500 = 29430 m²/s²

v_y = √29430  = 171.55 m/s

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The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

tan \theta = \dfrac{v_y}{v_x}  = \dfrac{171.55}{39.0 } = 4.4

∴ θ = tan⁻¹(4.4) = 77.19°.

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Explanation:

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Ksenya-84 [330]

Answer:

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Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

33\frac{1}{3}\ \frac{rev}{min}

Convert mixed number to an improper fraction

33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}

Remember that

1\ min=60\ sec

Convert rev/min to rev/sec

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Simplify

\frac{5}{9}\ \frac{rev}{sec}

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

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\frac{9}{5}\ \frac{sec}{rev}

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