Two cars are moving with velocities 70km/hr and 50km/hr in east and west direction respectively.

Physics

1 answer:

viva [34]3 years ago

5 0

Answer:

70 + 50 = 120 km/hr

Explanation:

The driver of either car would see the other car approaching or departing at 120 kph

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

ahrayia [7]

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.

3 0

3 years ago

The planet Gallifrey has 2 times the gravitational field strength and 2 times the radius of the Earth.

Alexeev081 [22]

The mass of the planet Gallifrey is 8 times the mass of the Earth.

let the gravitational field of Earth = g
let the radius of the Earth = R
gravitational field of Gallifrey = 2g
radius of Gallifrey = 2R

<h3>What is gravitational potential energy?</h3>

This is the work done in moving an object to a certain distance against gravitational field.

The gravitational field strength of the Earth is given as follows;

$g = \frac{GM}{r^2} \\\\G = \frac{gr^2}{M}$

The gravitational field strength of the Planet Gallifrey is calculated as follows;

$g_2 = \frac{GM_2}{r_2^2}$

$G = \frac{g_2r_2^2}{M_2} \\\\\frac{g_2r_2^2}{M_2} = \frac{gr^2}{M}\\\\M_2gr^2 = Mg_2r_2^2\\\\M_2 = \frac{Mg_2r_2^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times (2r)^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times 4r^2}{gr^2} \\\\M_2 = 8M$