Answer:
3.51s
Explanation:
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Answer:
a) 4.31 m/s²
b) 215.5 m
Explanation:
a) According to Newton's first law of motion
The net force applied to particular mass produced acceleration, a, according to
F = ma
F = 140 N
m = 32.5 kg
a = ?
140 = 32.5 × a
a = 140/32.5 = 4.31 m/s²
b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s
u = initial velocity of the probe = 0 m/s (since it was initially at rest)
a = 4.31 m/s²
t = 10 s
s = distance travelled = ?
s = ut + at²/2
s = 0 + (4.31×10²)/2 = 215.5 m
Im not 100% sure you have to tell me if im wrong or not.
D
B
C
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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