ΔU =
-Wint
Consdier the work of of
interaction is W =m*g*h - equation -1
and the Potential energy U.
Final Potential energy Uf =0
, And the Initial Potential Energy Ui =m*g*h
<span>Now we will write the
equation for a Change in Potential energy ΔU,</span>
ΔU = Uf
- Ui
= 0-m*g*h
<span> ΔU = -m*g*h --Equation 2</span>
Now compare the both equation
<span>Wint = -ΔU</span>
we can rewrite the above
equation
ΔU =
-W.
<span>So our Answer is ΔU = -W. .</span>
<span> </span>
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
The answer would be D) The tropical convection region because the states below SC are all tropical and the current would go up to SC. I hope this helps.
Answer: the photograph will likely show only one star.
Explanation:
Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.
Answer:
I think the answer is C too