During the 3.7 min a 7.4 A current is set up in a wire, how many (a) coulombs and (b) electrons pass through any cross section of the wire's width?

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.

Explanation:

Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).

Then, the intensity of electric current is expressed as:

Where:

In this case:

Replacing:

Solving:

Q= 7.4 A* 222 s

Q= 1,642.8 C

A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?

electrons= 1.39*10²¹

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s