Answer:
Find two boxes just a bit smaller than the other. The long sides of the box should be less than twice as long as the short side of the boxes. The smaller box should fit inside the larger box with about 1 inch in each direction to spare. The boxes can be cut down so that they fit together properly. Leave the flaps on the boxes. Buy a small sheet of Plexiglas (tm) a little bit smaller than the width and length of the top of the box. You will also need four pieces of cardboard to use for reflectors.
Explanation:
Answer:
<em>The total potential (magnitude only) is 11045.45 V</em>
Explanation:
<u>Electric Potential
</u>
The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

The total electric potential in A is


The total potential (magnitude only) is 11045.45 V
Answer:
The drag coefficient is
Explanation:
From the question we are told that
The density of air is 
The diameter of bottom part is
The power trend-line equation is mathematically represented as

let assume that the velocity is 20 m/s
Then


The drag coefficient is mathematically represented as

Where
is the drag force
is the density of the fluid
is the flow velocity
A is the area which mathematically evaluated as

substituting values


Then

Answer:
joules
joules
Explanation:
Let us convert the time in hours into seconds

Change in internal energy

where E is the internal energy in Joules
p is the power in watts
and t is the time in seconds

Joules
Amount of work done by the system

where P is the pressure and V is the volume
Substituting the given values in above equation, we get -

liter-atmospheres
Work done in Joules

Joules

Substituting the given values we get -

Thus
joules
joules