When fuel and air are compressed in the compression stroke, ...... a. each molecule of fuel is heated to its flash point b. each
1 answer:
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Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced
(a) The final speeds of the ice sleds is approximately 0.49 m/s each
(b) The impulse on the cat is 11.0715 kg·m/s
(c) The average force on the right sled is 922.625 N
The reason for arriving at the above values is as follows:
The given parameters are;
The masses of the two ice sleds, m₁ = m₂ = 22.7 kg
The initial speed of the ice, v₁ = v₂ = 0
The mass of the cat, m₃ = 3.63 kg
The initial speed of the cat, v₃ = 0
The horizontal speed of the cat, v₃ = 3.05 m/s
(a) The required parameter:
The final speed of the two sleds
For the first jump to the right, we have;
By the law of conservation of momentum
Initial momentum = Final momentum
∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'
Where;
v₁' = The final velocity of the ice sled on the left
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05
∴ 22.7 × v₁' = -3.63 × 3.05
v₁' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
For the second jump to the left, we have;
By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'
Where;
v₂' = The final velocity of the ice sled on the right
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05
∴ 22.7 × v₂' = -3.63 × 3.05
v₂' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
(b) The required parameter;
The impulse of the force
The impulse on the cat = Mass of the cat × Change in velocity
The change in velocity, Δv = Initial velocity - Final velocity
Where;
The initial velocity = The velocity of the cat before it lands = 3.05 m/s
The final velocity = The velocity of the cat after coming to rest =
∴ Δv = 3.05 m/s - 0 = 3.05 m/s
The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s
(c) The required information
The average velocity
Impulse = × Δt
Where;
Δt = The time of collision = The time it takes the cat to finish landing = 12 ms
12 ms = 12/1000 s = 0.012 s
We get;
∴
The average force on the right sled applied by the cat while landing, = 922.625 N
Learn more about conservation of momentum here:
Answer:
The percentage of its mechanical energy does the ball lose with each bounce is 23 %
Explanation:
Given data,
The tennis ball is released from the height, h = 4 m
After the third bounce it reaches height, h' = 183 cm
= 1.83 m
The total mechanical energy of the ball is equal to its maximum P.E
E = mgh
= 4 mg
At height h', the P.E becomes
E' = mgh'
= 1.83 mg
The percentage of change in energy the ball retains to its original energy,
ΔE % = 45 %
The ball retains only the 45% of its original energy after 3 bounces.
Therefore, the energy retains in each bounce is
∛ (0.45) = 0.77
The ball retains only the 77% of its original energy.
The energy lost to the floor is,
E = 100 - 77
= 23 %
Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %