Because it generates enough momentum to keep the train going with out really having to speed up
Answer:
Part a)

Part b)

Explanation:
As we know that magnetic flux through the loop is given as

now we have

now rate of change in flux is given as

now we know that



Now plug in all data


Part b)
Now the radius of the loop after t = 1 s



Now plug in data in above equation


The independent variable is the type of fuel used and the dependent variable is the speed of the race car. The independent variable could be changed through the experimental process to see its relation with the dependent variable<span>. The dependent variable is the result of the independent variable changes.</span>
Answer:
Q = 12540 J
Explanation:
It is given that,
Mass of water, m = 50 mL = 50 g
It is heated from 0 degrees Celsius to 60 degrees Celsius.
We need to find the energy required to heat the water. The formula use to find it as follows :

Where c is the specific heat of water, c = 4.18 J/g°C
Put all the values,

So, 12540 J of energy is used to heat the water.
Volocity can be the wave length of the speed like the volume.