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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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4 0
3 years ago
Pick one :))) hehe please
Arte-miy333 [17]

thermal energy is the correct answer of number 7



8 0
3 years ago
Read 2 more answers
On a straight road with the +x axis chosen to point in the direction of motion, you drive for 5 hours at a constant 20 miles per
Leviafan [203]

Answer:

v = 26.7 mph

Explanation:

During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:

Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi

Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:

Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi

So, the total displacement during all the trip wil be:

Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi

So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:

v = Δx / Δt = 160 mi / 6 h = 26.7 mph

8 0
3 years ago
SHOW ADEQUATE WORKINGS IN THIS SECTION
cluponka [151]

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

3 0
3 years ago
Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic
miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let X_A be its distance. In t hours he has walked

X_A=5t\ \text{miles}

Office B is walking at 6 mph directly west. In t hours his distance is

X_B=6t\ \text{miles}

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

D=\sqrt{X_A^2+X_B^2}

D=\sqrt{(5t)^2+(6t)^2}

D=\sqrt{61}t

This distance is known to be 2 miles, so

\sqrt{61}t=2

t =\frac{2}{\sqrt{61}}=0.256\ hours

t is approximately 15 minutes

3 0
3 years ago
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