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 magnitude of the force that each wire exerts on the other will increase by a factor of two.
Explanation:
force on parallel current carrying wire, F = BILsinθ
where;
B is the strength of the magnetic field
L is the length of the wire
I is the magnitude of current on the wire
θ is the angle of inclination of the wire
Assuming B, L and θ is constant, then F ∝ I
F = kI
When the amount of current is doubled in one of the wires, lets say the second wire;
Also, if will double the amount of current on the first wire, then
F₁ = 2F₂
Therefore, the magnitude of the force that each wire exerts on the other will increase by a factor of two.
Answer:
beneath the surface of the Pacific Ocean comes from samples and video collected by an unmanned lander
