Answer:

Explanation:
The expression which represent the first diffraction minima by a circular aperture is given by
--------eqn 1
The angle through which the first minima is diffracted is given by
---------eqn 2
As
is very small so we can write 
So from eqn 1 and eqn 2 we can write
--------eqn 3
Here
is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture
It is given that diameter of circular aperture is 14.7 cm so 
Now putting all these value in eqn 3


your answer should be “How does the number of radioactive atoms change over time?”
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We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y
Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m
displacement = √(116² + 333²)
= 353 m
Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.
Answer:
<em>Gravity</em><em>.</em><em> </em><em>The</em><em> </em><em>weight-force</em><em> </em><em>or</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>an</em><em> </em><em>object</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>because</em><em> </em><em>of</em><em> </em><em>Gravity</em><em>,</em><em> </em><em>which</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>object</em><em> </em><em>attracting</em><em> </em><em>it</em><em> </em><em>towards</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>earth</em><em>.</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x</em>
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