Answer:
I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).
The moment of inertia about the center of a sphere is 2 / 5 M R^2.
By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2
I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2
Let k = the force constant of the spring (N/m).
The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
= 0.5*(k N/m)*(0.35 m)²
= 0.06125k J
The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
= 0.5*(1.8 kg)*(1.9 m/s)²
= 3.249 J
Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m
Answer: 53 N/m (nearest integer)
Answer:
The mass of the beam is 0.074 kg
Explanation:
Given;
length of the uniform bar, = 1m = 100 cm
Set up this system with the given mass and support;
0-----------------33cm-----------------------------------100cm
↓ Δ ↓
0.15kg m
Where;
m is mass of the uniform bar
Apply the principle of moment to determine the value of "m"
sum of anticlockwise moment = sum of clockwise moment
0.15kg(33 - 0) = m(100 - 33)
0.15(33) = m(67)
Therefore, the mass of the beam is 0.074 kg
Answer:
Passed into the power grid for others to use the electricity
Explanation:
If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually passed into the power grid for others to use the electricity, generating a income to the homeowner
