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3 years ago

How does temperature change in layers of the atmosphere?

Physics

1 answer:

Gemiola [76]3 years ago

4 0

Answer:

Explanation:

The troposphere is hotter near the Earth's surface because heat from the Earth warms this air. ... Mesosphere: As the altitude increases, the air temperature decreases. The Mesosphere, like the troposphere layer, has a decrease in temperature with altitude because of the decreases in the density of the air molecules.

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Which best describes a difference between laser light and regular light?

Inga [223]

The basic difference is that the ordinary sources are incoherent that means that the discrete frequencies merge up to give an intermediate between the maximum and minimum frequencies. While the laser is coherent containing the single frequency with maximum amplitude. thus travelling far.

4 0

3 years ago

Plss help

Genrish500 [490]

Answer:

Explanation:

imagine urself on an elevator dont you feel lighter

3 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:

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:

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

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)

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.

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.

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

4 years ago

If wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:

GenaCL600 [577]

The frequency of the wheel is given by:
$f= \frac{N}{t}$
where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:
$f= \frac{100}{10 s}=10 s^{-1}$

And now we can find the angular speed of the wheel, which is related to the frequency by:
$\omega=2 \pi f=2 \pi (10 s^{-1})=62.8 s^{-1}$

6 0

4 years ago

Does this look like a good circuit diagram?

AnnyKZ [126]

It's hard to tell what's going on down there in the corner with the resistor and the ammeter. There seems to be as many as 3 or 4 wires in and out of the ammeter, which would be wrong. A real ammeter only has two ... one in and one out. (Same for a resistor.)

It's hard to say whether this circuit works, until we can clearly understand how everything is hooked up in that corner of the drawing.

6 0

3 years ago