Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :

Hence, the required work done is 1786.17 J.
Similarity : inverse square law for strength of force compared with distance.
<h2>Hey there!</h2>
The Force "F" applied on the unit electric charge "q" at a point describes the electric field.
<h3>☆ Formula to find electric charge:</h3>
<h2>Hope it helps </h2>
Answer:

Explanation:
Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

Due to the definition of cross product, the magnitude of the torque is given by:

Where
is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when
is equal to one, solving for r:
