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34kurt
4 years ago
5

The electric potential difference of a 150 µF capacitor is measured across the terminals of the capacitor and found to be 5 volt

s. What is the potential energy of the capacitor?
Physics
2 answers:
Assoli18 [71]4 years ago
7 0

Answer:

it's 2 on edge

Explanation:

Lapatulllka [165]4 years ago
6 0
The potential energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
C is the capacitance of the capacitor
V is the potential difference across the plates of the capacitor

In our problem,
C=150 \mu F= 150 \cdot 10^{-6} F
V=5 V
Therefore, the potential energy of the capacitor is
U= \frac{1}{2}(150 \cdot 10^{-6}F)(5 V)^2 = 1.88 \cdot 10^{-3}J
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Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would
Tatiana [17]

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight P_{1} = 30 W

Power of starter P_{2} = 2500 W

Voltage of headlight and starter V = 12 V

From equation of power,

 P = \frac{V^{2} }{R}

 R = \frac{V^{2} }{P}

For finding the resistance of headlight and starter,

⇒ For headlight,

 R_{1}  = \frac{144}{30} = 4.8 Ω

⇒ For starter,

R_{2} = \frac{144}{2500} = 0.057 Ω

Since equivalent resistance,

R_{eq} = R_{1} + R_{2} + ........

R_{eq} = 4.8 +0.057 = 4.857 Ω

So power in series is given by,

 P_{s } = \frac{V^{2} }{R_{eq} }  = \frac{144}{4.857}

 P_{s } = 29.64 W

8 0
4 years ago
Why do astronomers think that the milky way is a spiral galaxy?
Nikolay [14]

1) When you look toward the Galactic Center with your eye, you see a long, thin strip. This suggests a disk seen edge-on, rather than a ellipsoid or another shape. We can also detect the bulge at the center. Since we see spiral galaxies which are disks with central bulges, this is a bit of a top.

5 0
3 years ago
Taste and smell are called chemical senses because:
Bingel [31]

Answer:

The answer is a," their receptors are sensitive to chemical molecules."

3 0
4 years ago
A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6 * 1016 m, awa
igomit [66]

Answer:

At a distance of 1376.49 candle emits 0.2 watt power

Explanation:

Distance between Sun and earth 6.6\times 10^{16}m

Sun emits a power of P=3.8\times 10^{26}watt

Power emitted by candle = 0.20 watt

We know that brightness is given by

B=\frac{P}{4\pi d^2}

So \frac{3.8\times 10^{26}}{4\pi (6\times 10^{16})^2}=\frac{0.20}{4\pi d^2}

3.8\times 10^{26}d^2=7.2\times 10^{32}

d^2=1.89\times 10^6

d=1376.49m

So at a distance of 1376.49 candle emits 0.2 watt power

3 0
3 years ago
Help please*_*
DochEvi [55]

Answer:

3.83×10⁻¹⁹ J

Explanation:

The energy of a photon can be found using the formula: E=hc/λ

  • h is Planck's Constant which is approximately 6.63×10⁻³⁴
  • c is the speed of light which we'll round to 3.00×10⁸
  • λ is the wavelength of the photon. You typed 5200 but that's not even within the visible light spectrum's wavelength range, so I'm going to assume you meant 520 nm; or 5.20×10⁻⁷ m.

Plugging all that info in should give approximately 3.83×10⁻¹⁹ Joules of energy.

(You could get slight variations depending on what you chose to round up or down or not at all.)

5 0
3 years ago
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