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2 years ago

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2. You race a child that is exactly half your mass up identical ladders. If you took 5 s, and the child took 7 s, who did more w

ork? Who had greater power?

1 answer:

topjm [15]2 years ago

5 0

Answer:

You did more work, and you also had greater power.

Explanation:

Work is defined as a force doing a motion.

Particularly, when we lift an object of mass M by a distance H, the work is written as:

W = M*g*H

where g is the gravitational acceleration, then M*g = weight.

In this case, you and the child both are at the same distance to the ground after both of you finish the race, but your mass is twice the mass of the child.

Then if the child's mass is M, then the child's work is:

Wc = M*g*H

your work will be:

W = (2*M)*g*H

meaning that your work is twice the work.

Power is defined as the amount of work you do in a given amount of time.

So, if you needed 5 seconds to do the work W = (2*M)*g*H, then the (average) power will be equal to the quotient between the total amount of work you did and the time it took to do it.

Then your work is:

P = W/5s = (2*M)*g*H/5s

And the power for the child will be:

Pc = Wc/7s = M*g*H/7s

Your power has a larger numerator and a smaller denominator then is easy to see that you had greater power.

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3 years ago

Monochromatic light of a given wavelength is incident on a metal surface. However, no photoelectrons are emitted. If electrons a

mrs_skeptik [129]

Answer:

Light of a shorter wavelength should be used.

Explanation:

This is studied in the phenomenon called photoelectric effect, in which light is able to release electrons from a metal, said electrons are called photoelectrons .

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And a higher frequency corresponds to a shorter wavelength according to the equation:

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3 years ago

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 $=0.33\ m$ .

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3 years ago