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kumpel [21]
3 years ago
15

Electronic flash unit cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash l

asts for 1/675 s with an average light power output of 2.7❝105 W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy) how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125V when the stored energy equals the value calculated in part (a). What is the capacitance?
Physics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

(A) 421 J energy stored in the capacitor for one flash.

(B) The value of capacitance is 0.0537 F

Explanation:

Given :

(A)

Time t = \frac{1}{675}

Average power P = 2.7 \times 10^{5} W

From power equation,

   P= \frac{E}{t}

So energy in one light is given by,

   E = Pt

  E = 2.7 \times 10^{5} \times \frac{1}{675}  = 400 J

Since efficiency is 95 % so we can write, energy stored in one flash,

E_{tot} = \frac{400}{0.95} = 421 J

(B)

From the formula of energy stored in capacitor,

 E = \frac{1}{2}C V^{2}

Where E = E_{tot} and V = 125 V

 C = \frac{2E}{V^{2} }

 C = \frac{2 \times 421}{15625}

 C = 0.0537 F

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Answer:

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P = W / t       power that must be input (in Watts)

P = m g h / t = 1500 kg * 9.8 m/s^2 * 360 m /  60 sec

P = 88,200 watts

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A train travels 65 kilometers in 5 hours, and then 61 kilometers in 4 hours. What is its average speed? _______km/hr
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The trains average speed is 13 km/hr.
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3 years ago
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The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size aa ba
tester [92]

Answer:

3 volts

Explanation:

Number of turns in primary coil = N_{p} = 100

Number of turns in secondary coil = N_{s} = 200

Voltage across primary coil = V_{p} = 1.5 volts

Voltage across secondary coil = V_{s} = ?

In a transformer, the ratio of number of turns of primary to secondary coil is equal to the ratio of the respective voltages i.e.

\frac{N_{p} }{N_{s} } =\frac{V_{p} }{V_{s} }\\

Using the given values, we get:

\frac{100}{200}=\frac{1.5}{V_{s} }\\V_{s}=1.5 \times \frac{200}{100}\\V_{s}=3

Thus, the voltage measure across secondary coil would be 3 volts

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Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water
artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

4 0
3 years ago
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