An iron bolt of mass 57.5 g hangs from a string 38.8 cm long. the top end the right from the previously vertical line f the string is fixed. without touching it, a magnet attracts the bolt so that it remains stationary, but is displaced horizontally 22.5 cm to the string (a) draw a free-body diagram f the bolt. (b) find the tension the string. (c) find the magnetic force on the bolt.

tension is 0.692 N

magnetic force is 0.401 N

Explanation:

given data

mass = 57.5 g = 0.0575 kg

long = 38.8 cm = 0.388 m

displace = 22.5 cm = 0.225 m

to find out

free-body diagram ,  tension and magnetic force

solution

we consider here angle is θ

so free body diagram is attach below

and

here cosθ = x/l

so θ = cos^-1 ( 0.225 / 0.388)

θ = 54.6°

so tension from net  force is

net vertical force

t sinθ - F(g) = 0 ...........1

t = F(g) / sinθ

t = mg /sin54.6

t = 0.0575 ( 9.8) / sin54.6

t = 0.692

so tension is 0.692 N

and

net horizontal force

- t cosθ + F(magnetic) = 0   ....................2

t cosθ = F(magnetic)

put here all value

F(magnetic) = 0.692 × cos 54.6

F(magnetic) = 0.401

so magnetic force is 0.401 N

a displacement distance is size of displ

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.