It supports 128 primary partitions.
A group of protons and neutrons surrounded by electrons
Answer:
0.500 T
Explanation:
Since the change in time and the number of coils are both 1, I set the problem up to be 1.3=(1.5(x)-13(x)). I then plugged in numbers for x until I got the answer to be 1.3 V.
Answer:
Electric field acting on the electron is 127500 N/C.
Explanation:
It is given that,
Mass of an electron, 
Charge on electron, 
Initial speed of electron, u = 0
Final speed of electron, 
Distance covered, s = 2 cm = 0.02 m
We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :



According to Newton's law, force acting on the electron is given by :
F = ma


Electric force is given by :
F = q E, E = electric field


E = 127500 N/C
So, the electric field is 127500 N/C. Hence, this is the required solution.