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3 years ago

What layers on earth are considered liquids

1 answer:

6 0

Crust. The thin solid outermost layer of Earth. ...Asthenosphere. The lower layer of the crust. ...Lithosphere.Plasticity: is solid but still being able to. flow without being a liquid.The cool, rigid outermost layer of the Earth. ...the solid part of the earth consisting of the crust and outer mantle.

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"Compared to infrared radiation, does ultraviolet radiation have longer or shorter wavelengths? Does ultraviolet radiation have

balu736 [363]

Answer:Ultraviolet radiation has shorter wavelengths and higher energy than infrared radiation.

Explanation: Electromagnetic radiation radiations which have both electrical and magnetic properties,they can be transmitted through space or through a medium.

It includes Gamma radiation, infra-red, visible light, Ultraviolet radiation etc they occur with different wavelength, the lower the wavelength the higher the Energy dissipated per photon. According to their order of decreasing wavelength and increased energy they are classified as follows.

RADIO WAVE, MICRO WAVE, INFRA-RED, VISIBLE LIGHT, ULTRAVIOLET RAY, X-RAY, GAMMA RAYS.

5 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

3 years ago

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow- dryer is 11 Amps, and th

AURORKA [14]

Answer:

(a) 1320 W

(b) 480 W

(c) E':E ≈ 11:2

Explanation:

(a) Applying,

P' = VI'................. Equation 1

Where P' = Power of the blow-dryer, V = Voltage, I = current rating of the blow-dryer.

From the question,

Given: V = 120 V, I' = 11 A

Substitute these values into equation 1

P = (120×11)

P = 1320 W

(b) Similarly,

P = VI................... Equation 2

Where P = Power of the vacuum cleaner. I = current rating of the vacuum cleaner.

Also Given: I = 4 A,

Therefore

P = 4(120)

P = 480 W

(c)

E' = P'/t'............. Equation 3

E = P/t................ Equation 4

Where E' = Energy of the blow-dryer, t' = time of use of the blow-dryer, E = Energy of the vacuum cleaner, t = time of use of the vacuum cleaner

From the question,

Given: t' = 15 minutes = (15×60) = 900 seconds, t = 30 minutes = (30×60) = 1800 seconds

Substitute these values into equation 3 and 4

E' = 1320/900

E' = 1.47 J,

E = 480/1800

E = 0.267

Therefore,

E':E = 1.47:0.267

E':E ≈ 11:2

5 0

3 years ago

Which of these is an example of convection?

uranmaximum [27]

Which of these is an example of convection?

 A. an egg boiling in water 

B. an egg frying in a pan

C. an egg exposed to a flame

D. an egg warming under a light

5 0

3 years ago

Read 2 more answers

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

3 years ago