<span>553 ohms
The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is
Xc = 1/(2*pi*f*C)
where
Xc = Reactance in ohms
pi = 3.1415926535.....
f = frequency in hertz.
C = capacitance in farads.
I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance.
The 120 rad/s frequency is better known as 60 Hz.
Substitute known values into the formula.
Xc = 1/(2*pi* 60 * 0.00000480)
Xc = 1/0.001809557
Xc = 552.6213302
Rounding to 3 significant figures gives 553 ohms.</span>
Answer:
Electromagnetic waves do not require a medium in order to transport their energy. Mechanical waves are waves that require a medium in order to transport their energy from one location to another. ... Sound is a mechanical wave and cannot travel through a vacuum.
1. electrons
2. positive to negative
3. insulator
4. TRUE
5. closed circuit
6. TRUE
7. series
8. TRUE
9. v=ir
10. TRUE
Hope this helps! :)
The
sun is a ball of hot gases containing different kinds of elements at different
cores. It has a very high temperature that radiates all throughout the Milky
Way galaxy. The sun has three main parts; photosphere, chromospheres
and corona. The outer core of a star located at the chromospheres contains
mostly of hydrogen. Inside the hydrogen is helium then carbon, oxygen, neon,
magnesium silicon and the inert gas. The photosphere is scattered by the loose electrons in the corona’s plasma.
Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C = 
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C = 
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ = 
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
