An athlete at the gym holds a 1.5 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg .

a. What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?
b. What is the magnitude of the torque about his shoulder if he holds his arm straight, but 60 ∘ below horizontal?

torque τ   = 24 N-m

torque  τ  = 11.98 N-m  

Explanation:

given data

mass of the ball = 1.5 kg

arm length l = 70 cm = 0.7 m

mass of arm = 4 kg

solution

we get here  center of mass that is

center of mass =  ..........1

center of mass =  

center of mass = 0.44545 m  

and

torque will be

torque τ   =  F × r ×sinθ  ...........2

torque τ   =  (4+1.5)9.8 × 0.4454 × sin90

torque τ   = 24 N-m

and now we get magnitude of the torque  

magnitude of the torque  τ  = F × r ×sinθ

magnitude of the torque  τ  = (4+1.5)9.8 × 0.4454 × sin60

magnitude of the torque  τ  = 11.98 N-m  

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

answer: electrons