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Ahat [919]
2 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]2 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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3 years ago
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Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi
Anon25 [30]
Consider velocity to the right as positive.

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Second mass:
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v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
   = 4*2 + 8*(-3)
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Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
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(4+8 kg)*(v m/s) = -16 (kg-m)/s
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The center of mass is moving at 1.33 m/s to the left.
5 0
3 years ago
How much thermal energy is required to raise the temperature of 5 kg of water 50℃? The specific heat of water is 4168 J/kg*C.
sasho [114]

Answer/solution:

Given :

Mass =5kg

T 1 =20 C,T 2  =100 ∘C

ΔT=100−20=80 ∘C

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(kgK)

Q=5×4200×80

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3 0
2 years ago
what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...
Katarina [22]

(C)

Explanation:

E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})

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4 0
2 years ago
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kotykmax [81]

The statement “Impulse is a vector quantity” is true about Impulse.

Answer: Option B

<u>Explanation: </u>

The object’s action by applied force in a particular time interval, there happens changing in momentum called impulse. It is denoted by a symbol ‘J’ or ‘imp’ and expressed in a unit ‘Ns’. As impulse depends on the acted force, when a collision arises from front, behind or side, the force’s direction would be differed.

                   \text {Impulse }=\text {Force } \times \text {time}=\vec{F} \Delta t

So, from this option A is false as impulse is not a force but changing momentum. The unit is not Newton, it is Newton second (Ns). The force direction differs (impulse direction) for each cases of collision, so option D also false. Hence, option B seems to be correct. Vector quantity deals with both direction and magnitude and important in motion study.

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