To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m
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Answer:
(a). 12 plants
(b). 3171 $
Explanation:
(a)first convert units of 100 billion kWh/year into Watts(W)
also convert the units of 1000 MW into Watts(W)
1 billion = 10^9
1 year = 365*24 = 8760 hrs
so
100 billion kWh/year = 1
=
W
1000 MW = 
no. of plants =
= 11.4
So 12 plants required
(b)
savings = unit price*total units
= 
= 3170.9 =3171 $
Answer:
0.15694 m
Explanation:
m = Mass of block = 
v = Velocity of block = 11.2 m/s
k = Spring constant = 110 N/m
Here the kinetic energy of the fall and spring are conserved

The amplitude of the resulting simple harmonic motion is 0.15694 m