3 years ago

How much farther from the charge is the 2000 v equipotential surface than the 3000 v surface?

Physics

1 answer:

olganol [36]3 years ago

7 0

Answer:

r1 -r2 = 3.75cm

Explanation:

Check the attached file for the solution

You might be interested in

A car accelerates from rest at a rate of 6.50 m/s2. Determine thevelocity of the car at t = 4.00 s.

aksik [14]

Answer:

1.62 m/s2

Explanation:

6.50 divided by 4

3 0

2 years ago

A 500-n crate needs to be lifted 1 meter vertically in order to get it into the back of a pickup truck. what gives the crate a g

worty [1.4K]

The weight an weight of the truck

6 0

3 years ago

Read 2 more answers

John is hiking and notices a small stream of water flowing down the side of the mountain. What part of the water cycle is John o

Alexus [3.1K]

Answer: Runofff

Explanation: because it said that it is going down the side of the mountain

3 0

2 years ago

A sled with no initial velocity accelerates at a rate of 3.2 m/s2 down a hill. How long does it take the sled to go 15 m to the

grin007 [14]

S = ut + 1/2 at^2
a = 3.2 m/s^2
s = 15m
Find t
15 = 1/2(3.2)t^2
15 = 3.2t^2/2
30 = 3.2t^2
30/ 3.2 = 9.38
Square root of 9.38 = 3.06
It takes 3.06 seconds

3 0

3 years ago

A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

$\dfrac{1}{2}mV^2=mg(X-X_o)$

$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

$\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}$

$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

So we can say that

$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

6 0

3 years ago