<h2>
<em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em><em><u>S</u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em></h2>
<em>1) The relationship in between the electrical energy carriesd by the transmission wires and the amount of the heat loss in it is due to the reason that when the electricity is flown through the wires there are some resistance found in these wires which creates a disturbance in the efficient flow of electricty.Also we know that current have an heating effect when it is in motion as due to if a large amount or magnitude of electricity is flown through the transmission wires it will carry a larger heat effected and also due to the resistance is provided by the wires and so the process of heat loss takes place.</em>
<em>2)It is important to minimize current in transmission wires due to minimize the heat loss and resistance on flowing electric current to make the system more efficient </em>
<em><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u></em><em> 3)Given Resistance = 250 ohms </em>
<em>Electric potential = 150 volts </em>
<em>so we know Power = </em>
<em>volt^2/Resistance = </em>
<em>=</em><em>(150^2/250)(ohms/volts)</em>
<em>=</em><em>(22500/250)watt = 9</em><em>0</em><em> </em><em>w</em><em>a</em><em>t</em><em>t</em><em> </em>
<em>4)Heat energy (H) = Power(P)×Time(t)</em>
<em>4)Heat energy (H) = Power(P)×Time(t)= (90×2)joules = 180 joul</em><em>e</em><em>s</em>
<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em>
Answer:
The work done on the athlete is approximately 2.09 J
Explanation:
From the definition of the work done by a variable force:
and substituting with the function of our problem:
Answer:
dbdbdheh eewr h eahehwGFTT5Q3JFX
Explanation:
FJGXJDGTFJSRRXAGFEWFWDdQDE
'
Answer:
θ = 4.716 10⁻⁶ rad
Explanation:
In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.
We use the diffraction equation for a slit
a sin θ = m λ
The minimum occurs at m = 1
sin θ = λ / a
Since the angles in these systems are very small, we can approximate the sine to its angle in radians
θ = λ / a
The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number
θ = 1.22 λ / a
Let's calculate
θ = 1.22 518 10⁻⁹ / 13.4 10⁻²
θ = 4.716 10⁻⁶ rad