in the service producing sector
Answer: 55.52 *10^-6 C= 55.52 μC
Explanation: In order to solve this question we have to take into account the following expressions:
potential energy stired in a capacitor is given by:
U=Q^2/(2*C) where Q and C are the charge and capacitance of the capacitor.
then we have:
Q^2= 2*C*U=
C=εo*A/d where A and d are the area and separation of the parallel plates capacitor
Q^2=2*εo*A*U/d=2*8.85*10^-12*1.9*10^-5*11*10^3/(1.2*10^-3)=
=55.52 *10^-6C
Answer:
Option C
Explanation:
Given that
Motor force is 250 N
Force of friction is 750 N
Weight is 8500 N
And, the normal force is 8500 N
Now based on the above information
Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force
Because they are placed in different habitable zones
Protons copy the number equal