(a) We can find the current flowing between the walls by using Ohm's law:

where

is the potential difference and

is the resistance. Substituting these values, we get

(b) The total charge flowing between the walls is the product between the current and the time interval:

The problem says

, so the total charge is

The current consists of Na+ ions, each of them having a charge of

. To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
Answer:
Part a)

Part b)
Direction = upwards
Explanation:
When ball is dropped from height h = 4.0 m
then the speed of the ball just before it will strike the ground is given as



Now ball will rebound to height h = 2.00 m
so the velocity of ball just after it will rebound is given as



Part a)
Average acceleration is given as



Part B)
As we know that ball rebounds upwards after collision while before collision it is moving downwards
So the direction of the acceleration is vertically upwards
They all produce light and stuff
Explanation:
PRIMERO ENCUENTRAS EL PESO DEL PARACAIDISTA
= 65 kg(
) = 637 N
CON LA FÓRMULA DE LA ENERGÍA POTENCIAL
U = 637 N(6000 m - 3500 m) = 1592500 J
If you take a fluid (i.e. air or water) and heat it, the portion that is heated usually expands. The same mass takes up more volume and as a consequence the heated portion becomes less dense than the portion that is<span><span> not heated.</span> </span>