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
Answer:
B. change
Explanation:
In a chemical reaction, the bonds in the molecules break apart and form new ones, thus b is the correct answer because they change.
EC_1 + EP_1 = EC2 + EP_2
EC_2 = 0
EC_2 = EP_1 - EP_2
EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
Answer:
s = 1.7 m
Explanation:
from the question we are given the following:
Mass of package (m) = 5 kg
mass of the asteriod (M) = 7.6 x 10^{20} kg
radius = 8 x 10^5 m
velocity of package (v) = 170 m/s
spring constant (k) = 2.8 N/m
compression (s) = ?
Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore
• Ei = Ef
• Ei = energy in the spring + gravitational potential energy of the system
• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}
• Ef = kinetic energy of the object
• Ef = \frac{1}{2}mv^{2}
• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}
• s = 
s = 
s = 1.7 m
The answer is true. It tells me I have to have 20 characters.
