Physics, 06.04.2020 23:15 suiani8

Now click on "Speed" and increase the mass of the skater to the maximum mass and put her close to the top of the ramp, hit play and make a note of how high she goes up the other side. Now decrease the mass of the skater and repeat the experiment. What did you notice?

Answers: 3

Show answers

Answers

Answer from: luhmama

The speed of the skater is not depended on the mass of the skater.

Explanation:

A skater is moving in a U shaped ramp with some mass m and velocity v.

We are basically required to find out the relationship between speed of the skater and mass of the skater.

When the mass of the skater is small:

As you can notice in the attached diagram 1, the skater is placed at the top of the ramp with some height h from the horizontal ground. The speed of the skater at this point is zero.

We know that the energy due to the position of a body is potential energy therefore, the skater has maximum potential energy at this point given by

PE = mgh

As you can notice in the attached diagram 2, the skater is in motion now and it has maximum speed at the lowest point on the ramp.

The potential energy is being converted into kinetic energy therefore, the skater has maximum kinetic energy at this point given by

$KE = \frac{1}{2} mv^{2}$

When the mass of the skater is large:

As you can notice in the attached diagram 3, the skater is placed at the top of the ramp now with a larger mass. The speed of the skater at this point is zero.

The potential energy of the skater is now increased due to the increase in the mass since the mass and PE are directly proportional.

PE = mgh

As you can notice in the attached diagram 4, the skater is in motion and it has maximum speed at the lowest point on the ramp.

The potential energy is being converted into kinetic energy and similarly skater has now greater kinetic energy due to increase in the mass.

$KE = \frac{1}{2} mv^{2}$

Notice that the total energy remains constant due to the principle of conservation of energy.

PE = KE

$mgh = \frac{1}{2} mv^{2}$

$gh = \frac{1}{2} v^{2}$

$v^{2} = 2gh$

$v=\sqrt{2gh}$

As you can notice, the speed of the skater is not depended on the mass of the skater rather it depends on the gravitational acceleration and height of the ramp which are constant in the case therefore, skater with small mass and larger mass both will have the same speed.

On the other hand, the total energy of the skater would be greater in the case of larger mass and lesser in the case of smaller mass.

Now click on Speed and increase the mass of the skater to the maximum mass and put her close to th

Answer from: Quest

e) none of the above answers are correct

Answer from: Quest

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right

Another question on Physics

Physics, 21.06.2019 20:20

When many atoms are split in a chain reaction, a large explosion occurs. this is an example of what type of energy conservation

Answers: 2

Answer

Physics, 21.06.2019 23:30

Two technicians are discussing a resistance measurement between the can-h and can-l wires. technician a says this measurement should be done with the ignition switch in the "run" position. technician b states that a measurement of 0 ohms indicates an open in the network. which technician is correct?

Answers: 3

Answer

Physics, 22.06.2019 01:20

Acar moves a distance of 50.0 km west, followed by a distance of 64.9 km north what was the magnitude of the displacement of the car, in units of kilometers?

Answers: 2

Answer

Physics, 22.06.2019 05:20

Given the position vector of the particle r(t)=(t+1)i+(t^2−1)j+2t k, find the particle's velocity and acceleration vectors at t=1

Answers: 1

Answer

You know the right answer?

Now click on "Speed" and increase the mass of the skater to the maximum mass and put her close to th...

Questions