A student throws a 5.0-newton ball straight up.

1 answer:

3 0

If the ball weighs 5 newtons, then that's the force of gravity that
attracts it to the center of the Earth. The force only depends on
the mass of the ball and its distance from the center of the Earth.
The force of gravity doesn't care whether the ball is moving or at
rest. It doesn't change significantly until the ball moves to some
significant distance off the Earth's surface, which isn't happening
if it's tossed by a human being.

So we have a downward force of 5 newtons on the ball at all times.
In order to find the net force on it, we have to identify any other
forces besides gravity that are also acting on it.

If the ball were moving through the air, then there would be a force
of air resistance on it, acting opposite to its motion. But when it's
at its maximum height, in the process of turning around from rising
to falling, there's an instant when it's not moving through the air at all,
so gravity is the only force acting on it. At that moment, the net force
on the ball is 5.0 newtons downward.

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A bicycle travels at a velocity of 2.33 m/s, and has a displacement of -58.3 m. How much time did it take?

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Answer:

The time will be: 25.02 seconds

Explanation:

Given

A bicycle travels at a velocity of 2.33 m/s.

i.e.

v = 2.33 m/s

and

displacement = d = 58.3 m

As we know the formula to find the time

Time = Displacement/Velocity

= d/v

= 58.3/2.33

= 25.02 seconds

Therefore, the time will be: 25.02 seconds

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A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.

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Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work ( $W$ ) done on an object is equal to the change in its kinetic energy ( $\Delta K$ ):

$W=\Delta K = K_{f}-K_{i}$ (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation $K=\frac{I\omega^{2}}{2}$ for the kinetic energy so:

$\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2}$ (2)

with $\omega_{i}$ the initial angular velocity, $\omega_{f}$ the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is $I=MR^{2}$ , using those on (2)