I believe that the answer is <span>C. The atmosphere scatters blue light more than yellow light.</span>
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.
The force is repulsive.
Explanation: Please see the attachments below
24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²
Explanation:
It is known that relation between torque and angular acceleration is as follows.

and, I = 
So, 
= 4 


So, 
= 1 
as 
=
Hence, 

Thus, we can conclude that the new rotation is
times that of the first rotation rate.