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Rufina [12.5K]
3 years ago
11

In a particular metal, the mobility of the mobile electrons is 0.0033 (m/s)/(N/C). At a particular moment, the electric field ev

erywhere inside a cube of this metal is 0.033 N/C in the x direction. What is the average drift speed of the mobile electrons in the metal at this moment
Physics
1 answer:
Lapatulllka [165]3 years ago
6 0

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

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Answer:

The kinetic energy of the particle will be 12U₀

Explanation:

Given that,

A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.

Constant force = 12F

According to question,

The kinetic energy is

U_{0}=Fx....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

F_{r}=12F

When the particle moves from point B to point A then,

We need to calculate the kinetic energy

Using formula for kinetic energy

U=F_{r}x

Put the value of F_{r}

U=12Fx

Now, from equation (I)

U=12U_{0}

Hence, The kinetic energy of the particle will be 12U₀.

7 0
3 years ago
An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?​
snow_lady [41]
The wire vibrates back and forth between the poles of the magnet.
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7 0
2 years ago
a fast charged particle passes perpendicularly through a thin glass sheet of index of refraction 1.5. The particle emits light i
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The minimum speed of the particle is the Speed of light in glass is c/μ=2×108m/s.

<h3>Why is the refractive index important?</h3>

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7 0
1 year ago
Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00
MariettaO [177]

Answer:

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

d\times sin\theta=m\times \lambda

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

sin\theta=\frac {\lambda}{d}\times m ....1

The location of the bright fringe is determined by :

y=L\times tan\theta

Where, L is the distance between the slit and the screen.

For small angle , sin\theta=tan\theta

So,  

Formula becomes:

y=L\times sin\theta

Using 1, we get:

y=L\times \frac {\lambda}{d}\times m

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

0.0150\ m=2.00\ m\times \frac {518\times 10^{-9}\ m}{d}\times 1

<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>

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iVinArrow [24]
I believe it is A :) hope this helped
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3 years ago
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