What do it mean by the value of gravitational constant is 6.67×10^-11Nm^2/kg^2<br>

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.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

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

Describe the changes of motion when marble is moved from one viewpoint to another.

slavikrds [6]

Answer:

Explanation:

according to Newton First Law of Motion (Law of Inertia); An object at rest will stay at rest, forever, as long as nothing pushes or pulls on it. An object in motion will stay in motion, traveling in a straight line, forever, until something pushes or pulls on it.

the marble will move in a straight line

5 0

3 years ago

Moon problem please help!

Mariulka [41]

Answer:

crescent Moon crescent Moon

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3 years ago

Read 2 more answers

A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?

NemiM [27]

The average current is 0.10 A.

<h3>Current </h3>

Charge moving through a location on a circuit at a constant rate is called current. When numerous coulombs of charge pass over a wire's cross section in a circuit, it produces a large current. It is not necessary for a wire to be moving at a fast speed in order to have a high current if the charge carriers are tightly packed into the wire. To put it another way, many charge carriers traveling through the cross section is sufficient; they do not need to travel a great distance in a single instant. The amount of charges that flow through a cross section of wire on a circuit, as opposed to how far they travel in a second, is what determines current.

A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?

Learn more about current here:

brainly.com/question/2264542

#SPJ4

6 0

1 year ago

A circular loop of radius 0.0400 m is

Alenkinab [10]

Let's see

Find enclosed area of loop

$\\ \rm\dashrightarrow \pi r^2$