A basketball weighting 0.5kg is flying through the air at 10m/s. what is the kinetic energy

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

KE = Kinetic Energy and PE = Potential Energy

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

What is work?(theoretically)

emmasim [6.3K]

Answer:

force×distance

Explanation:

work is the ability of an object to move a distance as a result of the force being applied

4 0

3 years ago

how much horse power is needed to move a box that has a force of 1,492 newtons a distance of 2 meters in 1 second

stellarik [79]

This one is a piece 'o cake if you know the definition
of a horsepower, and impossible if you don't.

It is: 1 horsepower = 746 watts

Also, remember that 1 watt = 1 joule/second

Now: Work = (force) x (distance)

To move this box 2 meters,

   Work = (1,492 newtons) x (2 meters) = 2,984 joules

If you accomplished that feat in 1 second,
then you produced power of

   (2,984 / 1) (joule/second) = 2,984 watts .

   (2,984 watt) x (1 HP / 746 watt) = 4 horsepower .

That's the solution, applying bullet-proof math and physics to the
given data. But in the real world, I guarantee that you didn't do that.
______________________________________

By the way ... a comment regarding the terminology in the question:

The box doesn't "have a force" of 1,492 newtons. That's the force
with which you ... or a horse, or an ox ... pushed against the box
in order to move it.

(And, if I might observe, that force is about 336 pounds,
so my money would be on the horse or the ox.)

5 0

2 years ago

If the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 60 V, the current flowing through the circ

dem82 [27]

Question 2.

A transmission system at a radio station consists of essentially the following devices:

1. An oscillator, which converts a direct current into a high-frequency alternating current.

2. A transmitting antenna, which sends out radio waves having the same frequency as the current supplied by the oscillator.

3. A microphone, which picks up the sound waves and converts them into a suitable electric current with the same pattern as the sound waves.

4. A modulator, which enables the waves produced by the antenna to “carry” the pattern of the sound waves picked up by the microphone.

The answer is C. Oscillator

4 0

3 years ago

The mass of 29000 N elevator is

Lana71 [14]

In this case we are given a force of 29,000 N. We know that the equation we can use is:

Force = mass * acceleration

However this is not just any type of force, because this is the weight of the elevator. So when the force is equal to weight, the acceleration acting is simply gravity. Therefore:

weight = mass * gravity

Calculating for mass:

mass = 29,000 N / (9.8 m/s^2)

mass = 2,959.18 kg

5 0

3 years ago