Answer:
This question is incomplete
Explanation:
This question is incomplete because the telescope's focal length was not provided. The formula to be used here is
Magnification = telescope's focal length/eyepiece's focal length
The eyepiece's focal length was provided in the question as 0.38 m.
NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).
Answer:
option (b) 4900 N
Explanation:
m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R
F = G Me x m / (R + h)^2
F = G Me x m / 2R^2
F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2
F = 4900 N
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Answer:
Explanation:
Gravitational Potential Energy at earth surface 
Gravitational Potential Energy at height h is 
Energy required to lift the satellite 

Now Energy required to orbit around the earth



(given)




(b)For greater height
is greater than 
thus energy to lift the satellite is more than orbiting around earth