Complete question is;
Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.
What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?
Answer:
F = 501.5 N
Explanation:
We are given;
Mass of wooden crate; m = 75 kg
Angle of ramp; θ = 11°
Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;
F = mgsin θ + μmg cos θ
From online values, 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
Midway between the two<span> solstices we have equinoxes – Vernal Equinox in March and </span>Autumnal Equinox<span> in September. ... After this time, the Earth's northern axis is tilted </span>more<span> and </span>more<span>towards ... Then on </span>Summer Solstice<span>, the Sun will reach its farthest north position in the sky</span>
20/45=0.4*100= 44.4 so the answer is..................................................
Answer: 44.4%
