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2 years ago

What potential difference would an electron have to fall through to acquire a speed of 3.00*10^6 m/sec?

Physics

1 answer:

kirill [66]2 years ago

4 0

Answer:

25.6 V

Explanation:

The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.

m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C

eV = 1/2 m v^2

V = mv^2 / 2 e

V = (9.1 x 10^-31) x (3 x 10^6)^2 / (2 x 1.6 x 10^-19)

V = 25.6 V

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Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer

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Answer:

Explanation:

Let the angle between the first polariser and the second polariser axis is θ.

By using of law of Malus

(a)

Let the intensity of light coming out from the first polariser is I'

$I' = I_{0}Cos^{2}\theta$ .... (1)

Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.

By the law of Malus

$I'' = I'Cos^{2}\left ( 90-\theta \right )$

So,

$I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )$

$I'' = I_{0}Cos^{2}\theta Sin^{2}\theta$

$I'' = \frac{I_{0}}{4}Sin^{2}2\theta$

(b)

Now differentiate with respect to θ.

$I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta$

$I'' = \frac{I_{0}}{2}\times Sin 4\theta$

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2 years ago

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A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10

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Answer:

v = 895 m/s

Explanation:

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$T = 3369600 s$

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$r = 4.8 \times 10^8 m$

so the total path length is given as

$L = 2 \pi r$

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Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

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We know that

Acceleration= $\frac{v-u}{t}$

Using the formula

Acceleration= $a=\frac{0-4.5}{0.5}$

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Force =ma

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