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Zina [86]
2 years ago
7

How are a wave's energy and the wave's amplitude related?

Physics
1 answer:
dem82 [27]2 years ago
7 0
Answer:
Energy is proportional to the square of the amplitude

Explanation:
The energy of a certain wave is defined using its magnitude.
The two quantities are related directly. This means that as the amplitude of the wave increases, its energy increases and vice versa.

Energy is directly proportional to the square of the magnitude of the wave. This means that:
If we have new amplitude = 2 * old amplitude
We will have new energy = (2)² * old energy = 4 * old energy

Hope this helps :)
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Answer: When fire stopping material is used where more than 2 non-metallic sheathed cables pass through wood framing members, their ampacities must be adjusted, according to 310.15"

Answer is 2

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3 years ago
How do Air molecules enable sound to travel from a radios speaker to your ears?
MatroZZZ [7]
The radio frequencies push one air molecule that then bumps into a different air molecule.....which then hits another and another causing a line of crashing molecules that lead inside your ear and hits your ear drum causing it to vibrate which causes the sounds.
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Read 2 more answers
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
What does a trench form?What happens here?
Vinvika [58]
A hole, and people die if they fall in there.
8 0
3 years ago
Explain a situation in which you can accelerate even though your speed doesn’t change.
Serga [27]
"Acceleration" does NOT mean speeding up.  It also doesn't mean
slowing down.  Acceleration means ANY change in the speed
OR DIRECTION of motion.

The only kind of motion that's NOT accelerated is motion at a steady
speed AND in a straight line.

Even when your speed is steady, you're accelerating if your direction
is changing.

A few examples:
(no speeds are changing):

-- driving on a curved road, or turning a corner
-- going around a curve on a skateboard, a bike, or a Segway
-- running on a quarter-mile track
-- an Indy car cruising a practice lap around the track
-- water spinning, getting ready to go down the drain
-- any point on the blade of a fan
-- the little ball going around the inside of a Roulette wheel
-- the Moon in its orbit around the Earth
-- the Earth in its orbit around the sun
4 0
3 years ago
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