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Ahat [919]
3 years ago
15

2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and

Physics
1 answer:
Georgia [21]3 years ago
5 0

Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related.  When we graph P vs V, the area under the line is a trapezoid.  So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³.  So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa.  So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

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<h2>Answer:</h2>

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<h2>Explanation:</h2>

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Where;

M = sum of the masses

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m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

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M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

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