Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min
Answer: µ=0.205
Explanation:
The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,
f=Fw
The sum of the moments about the base of the ladder Is 0
ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²
Note that it doesn't matter WHAT the length of the ladder is -- it cancels.
Solve this for Fw.
0= 0.9637FwL - (67.91L)2.652
Fw=180.1/0.9637
Fw=186.87N
f=186.81N
Since Fw=f
We know Fw, so we know f.
But f = µ*Fn
where Fn is the normal force at the floor --
Fn = (25.8 + 67.08)kg * 9.8m/s² =
910.22N
so
µ = f / Fn
186.81/910.22
µ= 0.205
Can you get a better pic so i can read it
Pressure and heat. I hope this helps
A heliocentric system is a sun-centered