A cannonball and a marble roll smoothly from rest down an incline. Is the cannonball’s(a) time to the bottom and(b) translationa
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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
We are given information:
If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2
Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s
Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s
Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s
If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2
Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s
Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s
Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s