Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
Answer:
1%
Explanation:
Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.
The measured value is 2.02 meters and the actual value is 2.00 meters.
First, evaluate the fraction. Subtract 2.00 from 2.02
Next, divide 0.02 by 2.00
Finally, multiply 0.01 and 100.
The percent error is 1%.
Answer:
100/10 = 10 , 10 × 10 = 100÷20 = 5
I'm pretty sure its wrong
1km=1000m=1000000mm
118km/h=118000000 mm/h
Use stronger magnets
increase current
push magnets closer to coil
adding more sets of coils
