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3 years ago

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.

Physics

1 answer:

drek231 [11]3 years ago

8 0

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

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In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher

boyakko [2]

Answer:

L₀ = L_f , K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

L₀ = I₀ w₀

Final moment. Right after the crash

L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

L₀ = L_f

I₀ w₀ = (I₀ + mr²) w

w = $\frac{I_o}{I_o + m r^2}$ w₀

having the angular velocities we can calculate the kinetic energy

starting point. Before the crash

K₀ = ½ I₀ w₀²

final point. After the crash

K_f = ½ (I₀ + mr²) w²

sustitute

K_f = ½ (I₀ + mr²) ( $\frac{I_o}{I_o + m r^2}$ w₀)²

Kf = ½ $\frac{I_o^2}{ I_o + m r^2}$ w₀²

we look for the relationship between the kinetic energy

$\frac{K_f}{K_o}$ = $\frac{I_o}{I_o + m r^2}$

$\frac{K_f}{K_o } < 1$

K_f < K₀

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

6 0

2 years ago

A homeowner wants to reduce the environmental damage of the energy source used to heat her home.

liberstina [14]

Answer:

wind

Explanation:

wind is the only one of those onto that list that is a renewable source

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3 years ago

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