A 120 Ω resistor, a 60 Ω resistor, and a 40 Ω resistor are connected in parallel and placed across a potential difference of 12.
1 answer:
You might be interested in
To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.
The work done would be defined as
Where,
PE = Potential Energy
KE = Kinetic Energy
Where,
m = Mass
g = Gravitational energy
h = Height
v = Velocity
Considering power as the change of energy as a function of time we will then have to
The rate of mass flow is,
Where,
= Density of water
A = Area of the hose
The given radius is 0.83cm or m, so the Area would be
We have then that,
Final the power of the pump would be,
Therefore the power of the pump is 57.11W