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valentina_108 [34]
3 years ago
5

The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end

passes over a pulley and supports a 2-kg mass. The linear mass of the rope is 0.75kg/m. What is the wavelength of a transverse wave on the rope?
A. 0.25m
B. 0.16m
C. 0.12m
D. 1.2m
Physics
1 answer:
Darina [25.2K]3 years ago
5 0

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

T = mg

here we will have

T = 2(9.8) = 19.6 N

now we will have speed of wave given as

v = \sqrt{\frac{T}{\mu}}

here we will have

v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}

v = 16.33 m/s

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

\lambda = \frac{v}{F}

\lambda = \frac{16.33}{100} = 0.16 m

so wavelength will be 0.16 m

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At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon
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Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

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From the question we are told that

   The speed at point 1  is  v_1  =  3.57 \ m/s

   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

      h_2  =  h_1  -  18.5

Let the  diameter at point 1 be  d_1 then the diameter at point two will be

      d_2  =  2 * d_1

Now the continuity equation is mathematically represented as  

         A_1 v_1  =  A_2 v_2

Here A_1 , A_2  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e A=  \frac{\pi d^2}{4}]

   which can represent as

             A \ \  \alpha \ \  d^2

=>         A = c   d^2

where c is a constant

  so      \frac{A_1}{d_1^2}  =  \frac{A_2}{d_2^2}

=>          \frac{A_1}{d_1^2}  =  \frac{A_2}{4d_1^2}

=>        A_2  =  4 A_1

Now from the continuity equation

        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

=>     v_2  =  \frac{3.57}{4}

       v_2  =  0.893 \  m/s

Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

So  

         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

=>    P_2  =  \rho  * g  (h_1 -(h_1 -18.3)  + P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )

substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

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