Answer:
114.32195122 but Round your answer to three significant figures.) is 114
Explanation:
Just took the test
Here the block has two work done on it
1. Work done by gravity
2. Work done by friction force
So here it start from height "h" and then again raise to height hA after compressing the spring
So work done by the gravity is given as

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path


Total work done on it

So answer will be
None of these
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