A longitudinal wave is shown.<br> Which label identifies a rarefaction?

Applying the Law of Conservation of Energy. If a car was released down the track from a height what happens to the potential ene

erastova [34]

Answer:

According to the law of conservation of energy, energy cannot be created or destroyed, although it can be changed from one form to another. KE + PE = constant. A simple example involves a stationary car at the top of a hill. As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy. In the absence of friction, the car should end up at the same height as it started.

This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum. When the ball is at the top of its swing, all of the pendulum’s energy is potential energy. When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy. The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

4 0

2 years ago

Why must mine tailings be stored and disposed of carefully?

gavmur [86]

After thorough researching, the mine tailings must be stored and disposed of carefully because they have lots of chemical and various toxic materials. They can also leach to the aquifers. The correct answer to the following given statement above is they have chemicals which are dangerous.

7 0

3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

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

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

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

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

1.

Harlamova29_29 [7]

1.renewable energy
2.renewable energy
3.Turning off the lights when you are not in a room
4.conservation of energy
5.Driving an SUV
6.stored energy
7.Always from one side to another
8.use our energy resources wisely
9.conservation of energy
10.by either moving liquids or moving air currrents
11.potential energy
12.radiation
13.insulators
14.radiation
15.Evaporation
16.Thermal energy moves to an object with less heat.
17.Burning your finger on a hot match
18.potential energy
19.Energy is never being created or destroyed.
20.touching each other

7 0

3 years ago

Read 2 more answers

A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed o

Tju [1.3M]

Answer:

d. decreases

Explanation:

The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.

7 0

3 years ago

A longitudinal wave is shown. Which label identifies a rarefaction?