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
It is based on their state like population, literacy and poverty it varies from state to state, country to country
Answer:
B. Hold each type of fabric over a candle flame and time how long it takes for the fabric to start to burn.
Explanation:
Answer:
320 N/m
Explanation:
From Hooke's law, we deduce that
F=kx where F is applied force, k is spring constant and x is extension or compression of spring
Making k the subject of formula then
Conversion
1m equals to 100cm
Xm equals 25 cm
25/100=0.25 m
Substituting 80 N for F and 0.25m for x then
Therefore, the spring constant is equal to 320 N/m
