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Zepler [3.9K]
2 years ago
8

Find the energy u of the capacitor in terms of c and q by using the definition of capacitance and the formula for the energy in

a capacitor
Physics
1 answer:
gtnhenbr [62]2 years ago
3 0

The formula for the energy in a capacitor , u in terms of q and c is q²/2c

<h3>What is the energy of a capacitor?</h3>

The energy of a capacitor u = 1/2qv where

  • q = charge on capacitor and
  • v = voltage across capacitor.

<h3>What is the capacitance of a capacitor?</h3>

Also, the capacitance of a capacitor c = q/v where

  • q = charge on capacitor and
  • v = voltage across capacitor.

So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

= 1/2q(q/c)

= q²/2c

So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c

Learn more about energy in a capacitor here:

brainly.com/question/10705986

#SPJ12

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Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typica
kogti [31]

Answer:

The break even cost is $0.0063825

Explanation:

Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.

Wattage = W

Cost per kilo watt hour = C

Number of hours per year = H

Price per bulb/CFL = P

Discount rate = 11%

Life of bulb = 2 years

Price of bulb = $0.39

Wattage consumption of bulb per hours = 60

Life of CFL = 24 years

Price of CFL = $3.10

Wattage consumption of CFL per hour = 15

Calculate the Equated Annual Cost (EAC) of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)

PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods

PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)

<u>Calculate the EAC of bulb</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

EAC = {-0.39 – (51.37570 x C)}/ 1.712523, <em>consider this equation 1</em>

<u>Calculate the EAC of CFL</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, <em>consider this equation 2</em>

<u>Equate 1 and 2 to find the amount of C</u>

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

C = $0.0063825

Thus, the break- even cost per kilo – watt is $0.0063825

3 0
3 years ago
The heart working with the blood vessels to pump blood is which body system?
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I think it’s the cardiovascular system
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3 years ago
I NEED PHYSICS HELP ASAP! DO NOT GUESS! I WILL GIE BRAINLEIST IF RIGHT! THANKS! :D
ale4655 [162]
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An asteroid with a mass of 5.0 x 105 kg collides with the Earth and slides horizontally along the ground without bouncing (not a
V125BC [204]
The strength of the friction doesn't matter. Neither does the distance or the time the asteroid takes to stop. All that matters is that the asteroid has

1/2 (mass) (speed squared)

of kinetic energy when it lands, and zero when it stops.
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1/2 (mass) (original speed squared)

is the energy it loses to friction in order to come to rest.
8 0
3 years ago
The generator at a power plant produces AC at 20,000 V. A transformer steps this up to 355,000 V for transmission over power lin
Masja [62]

Answer:

Number of coil in the output is 39938

Explanation:

We have given a step up transformer

Input voltage of transformer, that is primary voltage v_p=20000volt

Output voltage, that is secondary voltage v_s=355000volt

Number of turns in primary N_p=2250

For transformer we know that \frac{V_p}{V_s}=\frac{N_p}{N_s}

\frac{20000}{355000}=\frac{2250}{N_s}

N_s=39937.5

As the number of turns can not be in fraction so number of turns in the output coil is 39938

7 0
3 years ago
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