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2 years ago

5

A scientist observes an increase in the rate at which water moves from the hydrosphere to the atmosphere at a particular locatio

n. Which of these conclusions about the location is most likely to be correct?

1 answer:

Musya8 [376]2 years ago

7 0

At a particular location, when an an increase in the rate at which water moves from the hydrosphere to the atmosphere, an increase in humidity is expected at that location. The term "humidity" generally refers to the amount of water vapor in the atmosphere.

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A long, hollow wire with inner radius a and outer radius b carries a uniform current density J. What is the magnetic field as a

sveta [45]

Answer:

The magnetic field in the region a < r < b is $B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

Explanation:

If we have the a < r < b. The formula of current is:

$J=\frac{I_{total} }{A}$

Where:

A = area enclosed by the loop.

Itotal = total current in loop.

$J=\frac{I}{\pi b^{2}-\pi a^{2} }$

$I_{enclosed} =JA_{enclosed}$

$I_{enclosed} =\frac{I(\pi r^{2}- \pi a^{2})}{\pi b^{2}-\pi a^{2} }$

If we have the Ampere`s law:

$\int\limits^a_b {B} \, ds =u_{0} I_{enclosed} \\2B\pi r=u_{0} (\frac{I(\pi r^{2}-\pi ^{2} }{\pi ^{2}-\pi a^{2} } )\\B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

6 0

2 years ago

Why is deforestation a serious global environment problem

julia-pushkina [17]

Answer:

we all know that deforestation is when tress are cut down but this may cause cardio dioxide to go up into the atmosphere and this may cause the rise in sea level and temperates tend to fluctuate

Explanation:

I'm just that smart yah dig

5 0

2 years ago

A radio technician measures the frequency of an AM radio transmitter. The frequency is . What is the frequency in megahertz? Wri

sukhopar [10]

Complete Question

A radio technician measures the frequency of an AM radio transmitter. The frequency is 14603 kHz . What is the frequency in megahertz? Write your answer as a decimal.

Answer:

The value is $x = 14.6 \ MHz$

Explanation:

From the question we are told that

The frequency is $f = 14603 \ kHz = 14603 *1000 = 14603000 \ Hz$

Generally

$1 Hz \to 1.0 *10^{-6} \ MHz$

$14603000 \ Hz \to x MHz$

=> $x = \frac{14603000 * 1.0*10^{-6}}{1 }$

=> $x = 14.6 \ MHz$

8 0

2 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass

vladimir2022 [97]

The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

6 0

2 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