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kolbaska11 [484]
2 years ago
7

Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral

lel to the first two. The total power dissipated in this circuit is: a. 2P/3 b. 3Pc. 2Pd. 3P/2e. P

Physics
1 answer:
Tamiku [17]2 years ago
6 0

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit  such that

P=VI=V^2/R

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

Current2=\frac{V}{2R}

Power2=Voltage * Current2

Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2

TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}

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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
Convert 172 to a scientific notation
gtnhenbr [62]

Answer:

1.72×10^2

Explanation:

Is the answer correct ?

6 0
2 years ago
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Elden [556K]

Answer:

D

Explanation:

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Volcanic eruptions are caused primarily by the movement of
evablogger [386]
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Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke
nirvana33 [79]

Answer:

4 times greater

Explanation:

<u>Step 1:</u> Calculate light-collecting area of a  20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² =\frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 20-m

A_{1} = \frac{\pi d^{2}}{4}

A_{1} = \frac{\pi (20^{2})}{4}

A₁ = 314.2 m²

<u>Step 2:</u> Calculate light-collecting area of a  10-meter Keck telescope (A₂)

A_{2} = \frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 10-m

A_{2} = \frac{\pi (10^{2})}{4}

A₂ = 78.55 m²

<u>Step 3</u>: divide A₁ by A₂  

= \frac{314.2 m^2}{78.55 m^2}

= 4

Therefor,  the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

5 0
3 years ago
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