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frez [133]
2 years ago
14

A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective dura

tion of the flash is 0.280 s, during which it produces an average 0.710 W from an average 3.00 V.
A) What energy does it dissipate?
B) How much charge moves through the lamp?
C) Find the capacitance.
D) What is the resistance of the lamp?
Physics
1 answer:
pshichka [43]2 years ago
8 0

Answer:

A. 0.199 J

B. 0.0663 C

C = 0.0221 F

D. 12.68 ohms

Explanation:

From the question:

time duration, t = 0.28 seconds

Average power, P = 0.71 W

Average voltage, V = 3 V

A) Energy is given as:

E = P * t

=> E = 0.71 * 0.28 = 0.199 J

B) Electrical energy is also given as:

E = qV

where q = charge

=> q = E / V

∴ q = 0.199 / 3 = 0.0663 C

C) Capacitance is given as charge over voltage:

C = q / V

=> C = 0.0663 / 3 = 0.0221 F

D) Electrical power, P, can also be given as:

P = V^2 / R

where R = resistance

=> R = V^2 / P

R = 3^2 / 0.71 = 9 / 0.71 = 12.68 ohms

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Why was Democritus considered a philosopher and not a scientist?
Alika [10]

Answer:

Democritus had no scientific instruments to extend the reach of his senses, so all of his experiments were just 'mind experiments', but because Democritus was a philosopher, he thought more into depth about why we humans are alive which led to the atomic theory.

Explanation:

5 0
3 years ago
What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

7 0
2 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
Which statement best describes why specific heat capacity is often more useful than heat capacity for scientists when comparing
maks197457 [2]

Answer:

Specific heat capacity is an intensive property and does not depend on sample size.

Explanation:

7 0
3 years ago
A ferry is travelling at 10 m/s relative to the river towards the south. The river is flowing at 2 m/s relative to
mixer [17]

Answer:

11 m/s south

Explanation:

The velocity of the passenger relative to the river bank is equal to the velocity of the passenger relative to the ferry, plus the velocity of the ferry relative to the river, plus the velocity of the river relative to the river bank.

v_passenger,bank = v_passenger,ferry + v_ferry,river + v_river,bank

If we take north to be positive and south to be negative:

v = 1.0 m/s + (-10 m/s) + (-2 m/s)

v = -11 m/s

v = 11 m/s south

8 0
3 years ago
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