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
Explanation:
The moment of inertia of each disk is:
Idisk = 1/2 MR²
Using parallel axis theorem, the moment of inertia of each rod is:
Irod = 1/2 mr² + m (R − r)²
The total moment of inertia is:
I = 2Idisk + 5Irod
I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]
I = MR² + 5/2 mr² + 5m (R − r)²
Plugging in values:
I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²
I = 23,750 g cm²
this can be solve using the formala of free fall
t = sqrt( 2y/ g)
where t is the time of fall
y is the height
g is the acceleration due to gravity
48.4 s = sqrt (2 (1.10e+02 m)/ g)
G = 0.0930 m/s2
The velocity at impact
V = sqrt(2gy)
= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)
V = 4.523 m/s
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