Explanation:
For a circular orbit v=
with G = 6.6742 × 
Given m = 6.42 x 10^23 kg and r=9.38 x 10^6 m
=> v = 2137.3 m/s
I hope this is the correct way to solve
Answer:
The natural medium emanating from the Sun and other very hot sources (now recognised as electromagnetic radiation with a wavelength of 400-750 nm), within which vision is possible.
Explanation:
just the way it is
Answer:
51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm
Explanation:
The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

Here,
= is the distance moved by the mirror M
is the wavelenght of the light used.
= 0.017m



Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7
Part a
Answer: NO
We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.
Using the second equation of motion:

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.
It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, 

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.
Part b
Answer: 29.6 m/s
The maximum distance that car can travel is 
The acceleration is same, 
The final velocity, v=0
Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.