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Ivan
3 years ago
6

A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative

velocity of 0.40c. What is the frequency of the beacon radio signal that is detected on the ship
Physics
1 answer:
Mekhanik [1.2K]3 years ago
8 0

Answer:

The frequency heard y the ship is 70 MHz.

Explanation:

True frequency, f = 50 MHz

speed of source, vs = 0

speed of observer, vo = 0.4 c

Let the frequency heard by the ship is f'.

Use the formula of doppler's effect

f' = \frac{v +v_o}{v+ v_s}f\\\\f' = \frac{c + 0.4 c}{c}\times 50\\\\f' = 1.4\times 50\\​\\f' = 70 MHz

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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
A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water a
Kaylis [27]

Answer:

Cp= 0.44 J/g.C

This is heat capacity of metal.

Explanation:

From energy conservation

Heat lost by metal = Heat gain by water +Heat gain by  calorimeter

Because here temperature of metal is high that is why it loose the heat.The temperature of water and  calorimeter is low that is why they gain the heat.

final temperature is T= 30.5 C

We know that sensible heat transfer given as

Q= m Cp ΔT

m=Mass

Cp=Specific heat capacity

ΔT=Temperature difference

By putting the values

55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)

Cp ( 99 .5- 30.5) = 30.65

Cp= 0.44 J/g.C

This is heat capacity of metal.

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A 10 kg bowling ball sits at the top of a 10 m hill and then slides down its icy hillside.
Nataly_w [17]
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A sound wave has a frequency of 247 Hz and a wavelength of 1.4 m. What is the speed of the sound wave in air?
zavuch27 [327]
The speed of sound through air at room temperature is almost always 343 m/s. However, since it doesn't tell you that, use the equation wavelength=velocity/frequency. Plug in the numbers: 1.4=v/247, and v=345.8 m/s. 
5 0
3 years ago
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