For both NPN and PNP this is true:
The base is between the collector and the emitter.
Achieve a full outer shell
Complete question:
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.
Answer:
The electric field inside this metal resistor is 3125 V/m
Explanation:
Given;
length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m
diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m
the potential difference due to electric field between the two ends of the resistor, V = 10 V
The electric field inside this metal resistor is given by;
ΔV = EL
where;
ΔV is change in electric potential
E = ΔV / L
E = 10 / (3.2 x 10⁻³ )
E = 3125 V/m
Therefore, the electric field inside this metal resistor is 3125 V/m
Hello!
momentum is p=mv so if the velocity is decreased by a factor of 2, which is 1/2 the value, then the momentum of that object will decrease by the same factorial.
Hope this helps, any questions please ask. Thank you!
Here in this case we can use work energy theorem
As per work energy theorem
Work done by all forces = Change in kinetic Energy of the object
Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.
Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

also we know that


Now kinetic energy is given by





Now by work energy theorem
Work done = 10500 - 0 = 10500 J
So in the above case work done on sphere is 10500 J