Define
v = volume of a drop per second, cm³/s
The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.
Therefore,
The average volume of a single drop is approximately 0.0556 cm³.
Answer: 0.0556 cm³
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
where
is the coefficient of friction, while N is the normal force. So we have:
since we know that F=300 N and
, we can find N, the magnitude of the normal force:
b) The problem is identical to that of the first part; however, this time the coefficienct of friction is
due to the presence of the oil. Therefore, we have:
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
