The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
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Answer:
The angle of incident ray is 40°.
Explanation:
Given that the angle of incident and reflected ray are the same. In this question, we had given that both angles added up will gives you 80° so you have to divide it by 2 :
incident + reflected = 80°
Let incident = reflected = θ
θ + θ = 80°
2θ = 80°
θ = 80° ÷ 2
= 40°
Answer:
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Explanation:
