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
When used as an energy source in a nuclear power plant, uranium is burned in a similar way as one would burn wood or coal for en
1 answer:
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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.