The full question reads;
Jason is employed at a moving company. A wooden crate weighing 75 kg is positioned on the wooden ramp of his truck, inclined at an angle of 11°.
What is the force magnitude, directed parallel to the ramp, that he needs to apply to initiate the upward movement of the crate?
Answer:
F = 501.5 N
Explanation:
We have the following information;
Mass of the wooden crate; m = 75 kg
Incline angle; θ = 11°
To move the wooden crate up, we must consider that friction is acting in the opposite direction of the movement along the inclined surface. Therefore, the force required can be expressed by;
F = mgsin θ + μmg cos θ
Using online resources, the coefficient of friction between wooden surfaces is μ = 0.5
Thus;
F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)
F = 501.5 N
Greetings!
Using the formula F = Bqv sin theta, we define F as Force, B as magnetic flux density, q as charge, v as velocity, and theta as the angle created by the moving electrons in relation to the magnetic field.
^^^You can compute the force using that equation^^^
In conclusion, your result would MOST LIKELY be "B".
"<span>-3.9 × 10-14 N"
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