Formula to find gravitational potential energy:
mgh
m: mass
g: gravitational acceleration
h: height (relative to reference level)
so the P.E. at 1.0.m is (5x9.8x1)= 49J
P.E. at 1.5m is (5x9.8x1.5) =73.5J
P.E. at 2.0m is (5x9.8x2)=98J
Answer:
Organic matter decomposition serves two functions for the microorganisms, providing energy for growth and suppling carbon for the formation of new cells. ... Dead plant residues and plant nutrients become food for the microbes in the soil
Explanation:
Answer:
0.37 m
Explanation:
Let the shoulder be the origin.
The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.
The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m
Substituting these into the equation for the center of mass, we have
y = (m₁y₁ + m₂y₂)/(m₁ + m₂)
y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)
y = (1.20 kgm + 0.65 kgm)/5.0 kg
y = 1.85 kgm/5.0 kg
y = 0.37 m
The distance of the center of mass from the shoulder is thus y = 0.37 m
Technically you can go forever on and on, but maybe your question was like how many rotations in a certain distance?
Answer:
The crate's coefficient of kinetic friction on the floor is 0.23.
Explanation:
Given that,
Mass of the crate, m = 300 kg
One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.
The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :
So, the crate's coefficient of kinetic friction on the floor is 0.23.
