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Scrat [10]
3 years ago
11

The duration of a photographic flash is related to an RC time constant, which is 0.200 µs for a certain camera. (a) If the resis

tance of the flash lamp is 0.0410 Ω during discharge, what is the size (in µF) of the capacitor supplying its energy? µF (b) What is the time constant (in s) for charging the capacitor, if the charging resistance is 650 kΩ? s
Physics
1 answer:
Alecsey [184]3 years ago
5 0

Explanation:

It is given that,

The time constant of RC circuit, \tau=0.2\ \mu s=0.2\times 10^{-6}\ s

(a) The resistance of the flash lamp is 0.0410 Ω

We know that the relation between the resistance and time constant is given by :

\tau=R\times C

C=\dfrac{\tau}{R}

C=\dfrac{0.2\times 10^{-6}}{0.0410}

C=4.87\times 10^{-6}\ F

(b) If he value of resistance, R=650\ k\Omega=650\times 10^3\ \Omega

\tau=R\times C

\tau=650\times 10^3\times 4.87\times 10^{-6}

\tau=3.16\ s

Hence, this is the required solution.

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Sound vibrations travel in a wave pattern, and we call these vibrations sound waves. Sound waves move by vibrating objects and these objects vibrate other surrounding objects, carrying the sound along. ... Sound can move through the air, water, or solids, as long as there are particles to bounce off of.

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(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

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<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
  • v is speed of sound

f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

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