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Harman [31]
3 years ago
9

I have a device that can generate sounds with frequencies between 800 Hz and 1600 Hz. I also have an unlabeled tuning fork that

I need to work out the fundamental frequency of. I find that it resonates at 920 Hz and 1380 Hz, but no other frequencies in that range. What's the lowest frequency that it will resonate at?
Physics
1 answer:
jek_recluse [69]3 years ago
5 0

Answer:

460 Hz

Explanation:

the given resonating frequency of the device

f₁ =  920 Hz and f₂ =  1380 Hz

fundamental frequency of the device is

f₀ = n₂ - n₁

   = 1380 - 920

   = 460 Hz

expression of frequency of organic pipe open at both ends

f_0=n\dfrac{\nu}{2l}

at n = 1

f_0=\dfrac{\nu}{2l} = 460 Hz

the frequency ratios of the closed pipe

f_0:f_1:f_2: ...... =[1:2:3:.........]f_0

                         =[1:2:3:.........]460 Hz

                         = 460 Hz : 920 Hz : 1380 Hz

so, the lowest frequency for the pipe open at both end is 460 Hz

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Complete Question

The  complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is   \Delta  P _1 =10  J/m^3

The pressure difference of the second bubble is  \Delta  P _2 =20  J/m^3

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

    The  radius of the first bubble is  r_1 =  10 \ mm=0.01 \ m

      The radius of the second bubble is  r_2  =  5 \ mm  =  0.005 \ m

      The surface tension of the soap solution is  s =  25 \ mJ/m^2 = 25*10^{-3} J/m^2

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

         \Delta  P  =  \frac{4 s}{R}

Now the pressure difference for the first bubble is  mathematically evaluated as

        \Delta  P _1 =  \frac{4 s}{r_1}

substituting values  

       \Delta  P _1 =  \frac{4 *25 *10^{-3}}{0.01}

       \Delta  P _1 =10  J/m^3

Now the pressure difference for the second bubble is  mathematically evaluated as

        \Delta  P _2 =  \frac{4 s}{r_1}

       \Delta  P _2 =  \frac{4 *25 *10^{-3}}{0.005}

       \Delta  P _2 =20  J/m^3

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A 2.0 kg cart has a momentum of 10.0 kg m/s. What is its velocity?
Svetach [21]

Answer:

\boxed {\boxed {\sf C. \ 5.0 \ m/s}}

Explanation:

Momentum is the product of mass and velocity. The formula is:

p=mv

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Substitute the values into the formula.

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3 0
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