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:
r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.
Explanation:
From a sum of forces:
where Ff = μ * N and 
N - m*g = 0 So, N = m*g. Replacing everything on the original equation:
(eq2)
Solving for r:

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.
Answer:
a) t = 1.47 h b) t = 1.32 h
Explanation:
a) In this problem the plane and the wind are in the same North-South direction, whereby the vector sum is reduced to the scalar sum (ordinary). Let's calculate the total speed
v =
f -
v = 585 -32.1
v = 552.9 km / h
We use the speed ratio in uniform motion
v = x / t
t = x / v
t = 815 /552.9
t = 1.47 h
b) We repeat the calculation, but this time the wind is going in the direction of the plane
v=
f -
v 585 + 32.1
v = 617.1 km / h
t = 815 /617.1
t = 1.32 h
there are 28 neutrons present
It would be D) the rope is pulled to the right. This is because their is a greater force in that direction.