Amuscle inserts 1.5 cm from the joint axis and exerts 300 n of force at an angle of pull of 60 degrees. how much torque is produced by this muscle? in your work, specify which element of the torque you manipulated.

torque = 3.897 N-m

Explanation:

given data

force = 300 N

angle = 60 degree

distance = 15 cm

to find out

torque

solution

we will apply here torque formula that is given below

torque = force × sinθ × distance    ...................1

put here all these value in equation 1

we get torque

torque  = force × sinθ × distance

torque  = 300 × sin60 × 1.5 ×

torque  = 300 × 0.8660 × 1.5 ×

torque  = 259.80 × 1.5 ×

torque  = 389.711 ×

torque = 3.897 N-m

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures