Acharged box (m=445 g, =+2.50 μc) is placed on a frictionless incline plane. another charged box =+75.0 μc) is fixed in place at the bottom of the incline. if the inclined plane makes an angle theta of 35.0∘ with the horizontal, what is the magnitude of the acceleration of the box when it is 59.0 cm from the bottom of the incline?

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

Where,

q = charges of the objects

r= distance/radius

Our values are previously given, so

Replacing,

The force acting on the block are given by,

Therefore the box is accelerated upward.

answer: the answer is a

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation: