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Murrr4er [49]
3 years ago
7

In one experiment the electric field is measured for points at distances r from a uniform line of charge that has charge per uni

t length λ and length l, where l≫r. In a second experiment the electric field is:

Physics
1 answer:
Yuki888 [10]3 years ago
8 0

Answer:

(a) A. Uniform line of charge and B. Uniformly charged sphere

(b) To three digits of precision:

λ = 1.50 * 10^-10 C/m

p = 2.81 * 10^-4 C/m^3

Explanation:

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B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

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v=\sqrt{\frac{2*e*V}{m} }    eq 1

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R=\frac{m*v}{q*B}       eq 2

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R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}

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Astar is 10 light years away from the earth. Suppose it brightens up suddenly today, after how long can we see this change?
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The phrase "light year" is a <u><em>distance</em></u> ... it's the distance that light travels through vacuum in one year.

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A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux
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C) 2.44 × 106 N/C

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The electric flux through a circular loop of wire is given by

\Phi = EA cos \theta

where

E is the electric field

A is the cross-sectional area

\theta is the angle between the direction of the electric field and the normal to A

The flux is maximum when \theta=0^{\circ}, so we are in this situation and therefore cos \theta =1, so we can write

\Phi = EA

Here we have:

\Phi = 7.50\cdot 10^5 N/m^2 C is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2

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3 years ago
What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K? The latent heat of vaporization
GREYUIT [131]

Answer : The mass of a sample of water is, 888.89 grams

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where,

q = heat = 2000 kJ = 2\times 10^6J       (1 kJ = 1000 J)

L = latent heat of vaporization of water = 2.25\times 10^6J/kg

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Now put all the given values in the above formula, we get:

2\times 10^6J=(2.25\times 10^6J/kg)\times m

m=0.88889kg=888.89g       (1 kg = 1000 g)

Therefore, the mass of a sample of water is, 888.89 grams

3 0
3 years ago
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