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:
0.67 s
Explanation:
This is a simple harmonic motion (SHM).
The displacement,
, of an SHM is given by

A is the amplitude and
is the angular frequency.
We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or
radian.
From trigonometry,
if A and B are complementary.
At
, 


So

At
, 





The period,
, is related to
by

Answer:
Yes
Explanation:
Testing electricity is a good conductor of electricity.