Answer:
the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².
Explanation:
Given;
applied force, F = 5000 N
cross sectional area of the steel wire, A = 0.2 m²
The tensile stress of the steel wire, is calculated as;
Therefore, the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².
Answer:
q = 8472.2 W/m
Explanation:
The solving or solution is given in the attach documents.
Until the object starts to move, the static coefficient of friction must be used since the object is at rest initially. If the horizontal force is large enough to overcome the friction force and the object accelerates, we switch to the kinetic coefficient.
The total downward force in the -y direction is the weight plus the applied 26N:
Fy=mg-26N = 5*-9.81-26 = -75.05N
The maximum friction force here is the magnitude of this force times the static coefficient of friction. This force will point in the opposite direction as an applied horizontal force, no matter which way it points (non conservative forces tend to oppose motion in all directions)
Ff=75.05N*1.03=77.3N
Note this is greater than the applied horizontal x-directed force of 26N. This means the answer is 26N of friction force acting on the object. Why not the full 77.3N? Because that is the maximum amount you can oppose along the x-direction before the object starts to move. But since we are only applying 26N, that is all the friction force pushes back with. This is a statement of Newton's 3rd Law of equal and opposite actions. If the object pushed back with the full 77.3N in this case it would accelerate backward (F=ma), since there would be a net force on the object (77.3N - 26N = 51.3N) This obviously doesn't happen!
Answer:
The minimum angle the ladder makes with the horizontal for the ladder not to slip and fall is 45 degrees.
Explanation:
Given that,
Mass of the ladder, m = 75 kg
Length of the ladder, l = 3 m
The center of gravity of the ladder is at a point 1.2 m from the base of the ladder.
The coefficient of static friction at the base of the ladder is 0.4
We need to find the minimum angle the ladder makes with the horizontal for the ladder not to slip and fall.
At equilibrium, net force and net torque on the ladder is equal to zero. Force is given by :
Using condition for torque as :
So, the minimum angle the ladder makes with the horizontal for the ladder not to slip and fall is 45 degrees.
Answer:
A. The sound wave will reflect off Buildings and automobiles.
Explanation:
This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.
