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Iteru [2.4K]
3 years ago
14

A stone is thrown into a well and the splash is heard after 3s. (speed of sound : 340m/s) . 1.) What time did the stone take to

reach the water? 2.) how long did it take for the sound to reach the observer? 3.) What is the depth of the well ?
P.s this question was super difficult for me even my physics teacher couldnt do it ( I know it's a shame) but I'm giving 50 points , so geeks start solving.
Physics
1 answer:
miss Akunina [59]3 years ago
8 0
1.)3s
2.)6s
3.)1020m

hope this help
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Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
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Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

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A = Area

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\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

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2 years ago
A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, its water level rises to 20.3 ml.
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Answer:

V=2.8 ml

Explanation:

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3 years ago
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tamaranim1 [39]
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A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input
KATRIN_1 [288]

Answer:

The peak-to-peak ripple voltage = 2V

Explanation:

120V and 60 Hz is the input of an unfiltered full-wave rectifier

Peak value of  output voltage = 15V

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dc output voltage = 14V

dc value of the output voltage of capacitor-input filter

where

V(dc value of output voltage) represent V₀

V(peak value of output voltage) represent V₁

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make C the subject of formula

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C = 2fR ((1 - (v₀/V₁))⁻¹

Substitute  for,

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C = 2 * 240 * 1 (( 1 - (14/15))⁻¹

C = 62.2μf

The peak-to-peak ripple voltage

= (1 / fRC)V₁

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