Answer:
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Answer:
The radius of the loop is 20.66 km
Explanation:
let the radius of the loop be r
mass of airplane is m
At the top, the pilot experiences two radial forces, which are
1) Gravitational force is mg
2) Centrifugal forces mv²/r out of the center
When the pilot experiences no weight,
then, mg = mv²/r
r = v² / g
= 450² / 9.8
= 20.66 x 10³3
= 20.66 km
Since the new distance is 3 times the old distance,
the new force is (1/3²) = 1/9th of the old force.
That's kind-of Choice-D, but I really don't like the way choice-D is worded.
"9 times smaller" is really pretty meaningless.
Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)
