PHYSICS COLLEGE

Find the average speed of the electrons in a 1.0 cm diameter, copper power line, when it carries a current of 20 A.

Answers

Answer 1

Answer:

Answer:

Average speed $v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec$

Explanation:

We have given current through power i = 20 A

Diameter d = 1 cm = 0.01 m

So radius r = 0.005 m

So area $A=\pi r^2=3.14* 0.005^2=7.8* 10^(-5)m^2$

Charge on electron e = $1.6* 10^(-19)C$

We know that current is given by $i=neAv_d$ , here n is nuber density of free electron, e is charge on electron, A is area and $v_d$ is average speed

We know that for copper n = $8.5* 10^(28)per\ m^3$

So average speed $v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec$

Related Questions

. An electron moving at 4.00×103m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40×10−16N . What angle does the velocity of the electron make with the magnetic field? There are two answers.
An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places
The International Space Station (ISS) is a space station orbiting the earth above the ground. If the radius of the earth is 3,958.8 miles, mass of earth is 5.972 x 10 24 kg, the period of the ISS at the orbit around the earth is 7.84 hours, can you calculate what is the distance from the ISS to the surface of the earth, in unit of miles
The unit of electric potential or electromotive force is the
What minimum distance would you have to hit a baseball from the center of the earth so that it would eventually reach the moon? Assume you can hit the ball directly along the line that connects the centers of the earth and moon. The distance between the centers of the earth and moon is ???? = 3.82 × 108 m.

a person sitting in a parked car hears an approaching ambulance siren at a frequency f1. as it passes him and moves away, he hears a frequency f2. the actual frequency f of the source is (which one of the following)a. f > f1b. f < f2c.f= f2 - f1d. f = f2 + f1e. f2 < f < f1

Answers

Answer:

e. f2 < f < f1

Explanation:

According to Doppler's Effect:

$(f_o)/(f_s) =(S+v_o)/(S-v_s)$ ......................................(1)

where:

$f_o\ \&\ f_s$ are observed frequency and source frequency respectively.

S = velocity of sound in the air from a stationary source

$v_o\ \&\ v_s$ are the velocity of the observer and the velocity of sound source with respect to a stationary frame of reference.

When the ambulance approaches a stationary observer

Here $v_o=0\$

Then eq. (1) becomes:

$(f)/(f1) =(S)/(S-v_s)$

Now, the value:

$(f1)/(f) =(S)/(S-v_s)>1$

$\therefore f<f1$

Now according to the given condition the source is moving away from the observer i.e. the velocity of the source is opposite to the velocity of sound with respect to the stationary observer.

Now the eq. (1) becomes

$(f2)/(f) =(S)/(S-(-v_s))$

∵the direction of motion of the source is away from the observer so a negative sign has been introduced.

Now, the value:

$(f2)/(f) =(S)/(S+v_s)<1$

$\therefore f>f2$

wo balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m. One is attached at one end of the rod and the other at the middle of the rod. If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,

Answers

Answer:

$T_1 =677224.40\ N$

Explanation:

given,

mass of the both ball = 5 Kg

length of rod = 2 L

where L = 0.55 m

angular speed = 45.6 rev/s

ω = 45.6 x 2 π

ω = 286.51 rad/s

v₁ = r₁ ω₁

v₁ =0.55 x 286.51 = 157.58 m/s

v₂ = r₂ ω₂

v₂ = 1.10 x 286.51 = 315.161 m/s

finding tension on the first half of the rod

r₁ = 0.55 r₂ = 2 x r₁ = 1.10

$T_1 = m ((v_1^2)/(r_1)+(v_2^2)/(r_2))$

$T_1 = 5 ((157.58^2)/(0.55_1)+(315.161^2)/(1.1))$

$T_1 =677224.40\ N$

What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.

Answers

Answer:

λ = -47 nC / m

Explanation:

The missing question is as follows:

" The potential difference between the surface of a 2.2 cm -diameter power line and a point 1.9 m distant is 3.8 kV. What is the magnitude of the line charge density on the power line? Express your answer using two significant figures. "

Given:

- The Diameter of the power line D = 2.2 cm

- The distance between two ends of power line L = 1.9m

- The potential difference across two ends V = 3.8 KV

Find:

What is the magnitude of the line charge density on the power line?

Solution:

- The derivation of the line of charges for a length L oriented along any axis centered at origin and the potential difference between two ends is as follows:

V = 2*k*λ*Ln( D / L )

Where,

k : Coulomb's Constant = 8.99*10^9

λ : The line charge density

- Re-arrange and solve for λ:

λ = V / 2*k*Ln( D / L )

Plug in the values:

λ = 3800 / 2*8.99*10^9*Ln( 2.2 / 190 )

λ = -4.74022*10^-8 C / m

λ = -47 nC / m

Final answer:

Line charge density is the total charge distributed along the length of a wire, expressed in coulombs per meter. To calculate it, divide the total charge by the total length of the wire. Without specific numbers for charge and length, a numerical value can't be given.

Explanation:

To calculate the magnitude of the line charge density of a power line, you need to know the total charge (Q) distributed along the total length (L) of the wire. The line charge density (λ) is then defined as λ = Q/L. Unfortunately, without any specific numbers provided for these parameters, I can't provide a numerical answer.

Line charge density is a significant concept in electromagnetism and is measured in coulombs per meter (C/m).

Remember that the charge can be uniform or non-uniform along the length of the line.

For example, if a power line has a total charge of 0.02 C spread along its length of 50 m, it would have a line charge density of λ = Q/L = 0.02 C / 50 m = 0.0004 C/m

Learn more about Line Charge Density here:

brainly.com/question/34815993

#SPJ3

A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m/s, what is the pin's final velocity?a 5 m/s
b 2.5 m/s
c 10 m/s
d 5.2 m/s

Answers

Answer:

The pin's final velocity is 5m/s

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

An alternating current is supplied to an electronic component with a rating that the voltage across it can never, even for an instant, exceed 16 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit?A)8 sqrt 2 V

B) 16 sqrt 2 V

C) 256 V

D) 8

Answers

Answer:

A) $V_(rms)=8√(2) V$

Explanation:

Maximum voltage = $V_(max)=16 V$

Maximum voltage and rms voltage are related to each other by

$V_(max)=V_(rms) * √(2) \nV_(rms)=(V_(max))/( √(2))\nV_(rms)=(16)/(√(2)) \nV_(rms)=8√(2) V$

While doing her crossfit workout, Yasmeen holds an 7.0 kg weight at arm's length, a distance of 0.57 m from her shoulder joint. What is the torque about her shoulder joint due to the weight if her arm is horizontal? A 30 N m B. 4.0 N m C. 43N-m D. 39 N m