how much tension (fm) must be supplied at the triceps to stabalize the arm against an external force (fe) of 200 N given dm

The missing part of the question is :

given dm-2 cm and de 25 cm

The correct answer is = 2500 N

Explanation:

Given:

External force (fe) = 200 N

dm- 2 cm = 0.02 m

de = 25 cm = 0.25 m

Balancing torque about elbow

Fm*dm = Fe*de

By putting value

fm*0.02 = 200*0.25

fm= 200*0.25/0.02

fm = 50/0.02

fm = 2500 N

Thus, the correct answer is = 2500 N

c.the charges if the proton and electron and gravity

6 cos 34, 6 sin 34 trig