A wrench is placed at 30 cm in front of a diverging lens with a focal length of magnitude 10 cm. What is the magnification of the wrench?

0.25

Explanation:

Magnification = image distance/object distance

mag = v/u Equation 1

Given: f = -10 cm ( diverging lens) u = 30 cm.

Where can calculate for the value of v using

1/f = 1/u+1/v

make v the subject of the equation

v = fv/(u-f) Equation 2

Substitute into equation 2

v = -30(10)/(30+10)

v = -300/40

v = -7.5 cm.

substituting into equation 1,

mag = 7.5/30

mag = 0.25

hence the magnification of the wretch = 0.25

grav potential energy

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards