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Amanda [17]
3 years ago
11

A baseball is travelling towards a player's bat with a speed of 40.0 m/s. After being hit by the bat, the baseball is travelling

wtih a speed of 45.0 m/s in the opposite direction of the initial velocity. Calculate the impulse imparted to the baseball.
Physics
1 answer:
laila [671]3 years ago
3 0
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a train has an initial velocity of 30 m/s. If the train accelerates uniformly at a rate of 6.3 m/s ^ for 2.8 seconds what is the
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Answer:

the velocity is a second final to initial velocity of 39

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Which describes acceleration involving only a change in direction?
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If we consider any system moving with u<span>niform circular motion we can notice that the MAGNITUDE of the accelaration remains constant. However,  there is a change in the direction of the acceleration at every instant of time .

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6 0
4 years ago
Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to t
Whitepunk [10]

Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).

This means the lungs are inflated with - Highly pressurized- gas.

This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.

If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.

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5 0
3 years ago
Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1
harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum p_{i} (before the elastic collision) must be equal to the final momentum p_{f} (after the elastic collision):

p_{i}=p_{f} (1)

Being:

p_{i}=m_{1}V_{i} + m_{2}U_{i}

p_{f}=m_{1}V_{f} + m_{2}U_{f}

Where:

m_{1}=200 kg +100 kg=300 kg is the combined mass of Tubby and Libby with the car

V_{i}=10 m/s is the velocity of Tubby and Libby with the car before the collision

m_{2}=25 kg + 100 kg=125 kg is the combined mass of Flubby with its car

U_{i}=0 m/s is the velocity of Flubby with the car before the collision

V_{f}=4.12 m/s is the velocity of Tubby and Libby with the car after the collision

U_{f} is the velocity of Flubby with the car after the collision

So, we have the following:

m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f} (2)

Finding U_{f}:

U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}} (3)

U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg} (4)

Finally:

U_{f}=14.1 m/s

8 0
3 years ago
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