First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency
Answer:

Explanation:
From the second law of Newton movement laws, we have:
, and we know that a is the acceleration, which definition is:
, so:

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

(This is the indefinite integral), the definite one is:

Answer :
.
Explanation:
It is given that,
Electric field strength, 
We know that,
Charge of electron, 
Mass of electron, 
From the definition of electric field,
...............(1)
According to Newton's second law, F = ma..........(2)
From equation (1) and (2)




or

So, the horizontal component of acceleration of an electron is
.
Hence, it is the required solution.
A decrease in it's operating temperature would make a heat engine less efficient. This is because in order to operate, a heat engine needs to be hot and maintain that temperature.