Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>
Two Factors That Affect How Much Gravity Is on an Object. Gravity is the force that gives weight to objects and causes them to fall to the ground when dropped. Two major factors, mass and distance, affect the strength of gravitational force on an object.
That's "<em><u>insolation</u></em>" ... not "insulation".
'Insolation' is simply the intensity of solar radiation over some area.
If 200 kW of radiation is shining on 300 m² of area, then the insolation is
(200 kW) / (300 m²) = <em>(666 and 2/3) watt/m²</em> .
Note that this is the intensity of the <em><u>incident</u></em> radiation. It doesn't say anything
about how much soaks in or how much bounces off.
Wait !
I just looked back at the choices, and realized that I didn't answer the question
at all. I have no idea what "1 sun" means. Forgive me. I have stolen your
points, and I am filled with remorse.
Wait again !
I found it, through literally several seconds of online research.
1 sun = 1 kW/m².
So 2/3 of a kW per m² = 2/3 of 1 sun
That's between 0.5 sun and 1.0 sun.
I feel better now, and plus, I learned something.
Answer:
the work converted to thermal energy is 40 J.
Explanation:
Given;
work done by the physicist,w = 100 J
height through which the book is raised, h = 0.2 m
efficiency of machine = 60% = 0.6
The useful work done by the machine is calculated as;
useful work = 0.6 x 100 = 60 J
The wasted energy = 100 J - 60 J
The wasted energy = 40 J
The wasted energy by the machine is possibly converted to thermal energy by the frictional part of the machine.
Therefore, the work converted to thermal energy is 40 J.
