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
Answer:

Explanation:
In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

In our problem, the work function of cesium is

so, we can convert it into Joules by using the following proportion:

Answer: Runofff
Explanation: because it said that it is going down the side of the mountain
Answer:
4 Ohms
Explanation
(This is seriously not as hard as it looks :)
You only need two types of calculations:
- replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two:

- replace two resistances that are connected in parallel. In that case:

I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the <em>equivalent resistance</em> of the original circuit.
Using these two transformations you will be able to solve step by step any problem like this, no matter how complex.