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jok3333 [9.3K]
3 years ago
9

Consider a long cylindrical charge distribution of radius R with a uniform charge density rho. Find the electric field at distan

ce r from the axis where r < R. (Use ε0, rho, R, and r as necessary.)
Physics
1 answer:
ahrayia [7]3 years ago
3 0

Answer: Ok, first lest see out problem.

It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.

Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0

where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.

So Q= rho*volume= pi*r*r*L*rho

so replacing : E = (1/2)*r*rho/ε0

you may ask, ¿why dont use R on the solution?

since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.

R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.

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A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Check all that
9966 [12]

Answer:

f.The period is independent of the suspended mass.

Explanation:

The period of a pendulum is given by

T=2 \pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that:

1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length

2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation

So, the only correct statements are

f.The period is independent of the suspended mass.

Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.

3 0
3 years ago
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
3 years ago
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Plz do it all i will give brainlest and thanks to best answer.
maxonik [38]
Its a, metal is a good conductor of heat so yea
Hope this helps :)
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