Answer:
The length of the tube is 3.92 m.
Explanation:
Given that,
Electric potential = 100 MV
Length = 4 m
Energy = 100 MeV
We need to calculate the value of 
Using formula of relativistic energy

Put the value into the formula


Here, 



We need to calculate the length
Using formula of length

Put the value into the formula


Hence, The length of the tube is 3.92 m.
Answer:
chimical change...or phisical...one or the other...
Explanation:
Answer:
9.8 secs
Explanation:
the ball is in the air so it takes 9.8 secs to get to the ground
The Answer is= 7.8 x 10^4
Answer:
Part a)
Velocity = 6.9 m/s
Part b)
Position = (3.6 m, 5.175 m)
Explanation:
Initial position of the particle is ORIGIN
also it initial speed is along +X direction given as

now the acceleration is given as

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero
so we will have



Part a)
the velocity of the particle at this moment in Y direction is given as



Part b)
X coordinate of the particle at this time



Y coordinate of the particle at this time



so position is given as (3.6 m, 5.175 m)