A95 n force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 n · m. what is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?

angle = 18.40 degree

Explanation:

given data

force = 95 N

distance = 0.50 m

torque = 15 N · m

to find out

angle between the wrench handle and the direction of the applied force

solution

we will apply here torque equation that is express as

torque =  distance × force × sin(θ)   1

put here value we will get angle that is

15 = 0.50 × 95 × sin(θ)

sin(θ) = 0.315789

θ = 18.40 degree

18.4°

Explanation:

Force, F = 95 N

Torque, τ = 15 Nm

length, r = 0.5 m

τ = r F Sinθ

where, θ be the angle between distance vector and force vector

So, 15 = 0.5 x 95 x Sinθ

Sin θ = 0.3159

θ = 18.4°

Thus, the angle is 18.4°.

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it has to be b, about 97 percent.