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spin [16.1K]
3 years ago
8

A piston-cylinder device initially contains 0.3 kg of nitrogen gas at 160 kPa and 140oC. The nitrogen is now expanded isothermal

ly to a pressure of 100 kPa. Determine the boundary work done during this process.
Physics
1 answer:
irakobra [83]3 years ago
4 0

Answer:

<h2>Work done by the gas is given as</h2><h2>W = 1.72 \times 10^4 J</h2>

Explanation:

As we know that the process is isothermal so here work done is given as

W = nRTln(\frac{P_1}{P_2})

here we know that

n = \frac{w}{M}

n = \frac{300}{28} = 10.7 moles

now we have

W = 10.7 (8.31)(273 + 140) ln(\frac{160}{100})

so we have

W = 1.72 \times 10^4 J

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A father uses a rope to pull a child on a sled at a constant speed. of the ones listed which forces are present?
Amanda [17]

All the forces may be present.

The child on the sled has a weight- which is due to the force of gravity on the child and the sled.

The sled and the child exert a force on the ground equal to the combined weight of the sled and the child. The ground exerts a normal force on the sled.

The force used by the father to pull the sled is the applied force.

The sled slides on the ground and as a result, the force of friction exists between the ground and the sled, directed opposite to the direction of the motion of the sled.

The father pulls the sled using a rope. As a result, the rope is under Tension.

As the sled moves it also experiences a force of air resistance, which is dependent on the sled's speed.

However, since the father pulls the sled along with constant speed, the sum of all the forces acting on the sled is zero.

Since there is no movement in the upward or downward directions, the weight of the child and the sled is equal to the normal force acted upon the sled by the ground.

The force applied by the father on the rope is equal to the tension in the rope.

Friction and air resistance act opposite to the direction of motion of the sled. for the sled to move at constant speed, the tension in the rope must be equal to the sum of the forces due to friction and air resistance.

4 0
3 years ago
Jack and Jill exercise in a 25.0 m long swimming pool. Jack swims 9 lengths of the pool in 156.9 s ( 2 min and 36.9 s ) , wherea
Leya [2.2K]

Answer:

Jill average velocity is  0

Jack  average velocity is 0.159337

Jill average speed = 1.593372

Jack average speed = 1.434034

Explanation:

given data

long swimming pool = 25.0 m

9 lengths of the pool = 156.9 s ( 2 min and 36.9 s )

10 lengths = same time interval

to find out

average velocity and average speed

solution

we know that average velocity that is express as

average velocity = \frac{displacement}{time}    .....................1

Jill come back where she start

so here velocity will be = 0

and

Jack ends up on the other end of pool

so average velocity =  \frac{25}{156.9}

average velocity = 0.159337

now we get here average speed that is express as

average speed = \frac{distance}{time}      .............2

jack speed = 9 × \frac{25}{156.9}

jack speed = 1.434034

and

Jill speed = 10 × \frac{25}{156.9}

Jill speed = 1.593372

5 0
3 years ago
Which statements explain the special theory of relativity? Check all that apply.
frozen [14]

Answer:

A. Time and space are relative.

B. The speed of light in a vacuum is constant for all observers.

D. Physical laws change based on an observer's motion.

E. Physical laws remain constant regardless of an observer's motion.

Explanation:

7 0
2 years ago
Read 2 more answers
Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t
Step2247 [10]

Answer:

\Delta p = 90.7 kPa

Explanation:

specific gravity of oil is = \frac{\rho_{oil}}{\rho_w}

\rho_{oil} = 0.85*1000 = 850 kg/m3

we know that

change in pressure  for oil is given as

\Delta p = \rho gh

here density and h is for oil

\Delta p = 850*5 *9.81 = 41,692.5 kPa

change in pressure  for WATER is given as

\Delta p = \rho gh

here density is for water and h is for water

\Delta p = 1000*5 *9.81 = 49,050 kPa

pressure change due to both is given as

\Delta p = 41692.3 + 49050 = 90742.5 N/m2

\Delta p = 90.7 kPa

8 0
3 years ago
Read 2 more answers
In an electromagnetics lab, you are studying two coils which have a mutual inductance of M=300 mH. Suppose that the current in t
Firdavs [7]

Answer:

Explanation:

Given that,

The mutual inductance of the two coils is

M = 300mH = 300 × 10^-3 H

M = 0.3 H

Current increase in the coil from 2.8A to 10A

∆I = I_2 - I_1 = 10 - 2.8

∆I = 7.2 A

Within the time 300ms

t = 300ms = 300 × 10^-3

t = 0.3s

Second Coil resistance

R_2 = 0.4 ohms

We want to find the current in the second coil,

The same induced EMF is in both coils, so let find the EMF,

From faradays law

ε = Mdi/dt

ε = M•∆I / ∆t

ε = 0.3 × 7.2 / 0.3

ε = 7.2 Volts

Now, this is the voltage across both coils,

Applying ohms law to the second coil, V=IR

ε = I_2•R_2

0.72 = I_2 • 0.4

I_2 = 0.72 / 0.4

I_2 = 1.8 Amps

The current in the second coil is 1.8A

7 0
3 years ago
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