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Artist 52 [7]
3 years ago
8

With what tension must a rope with length 2.20 m and mass 0.150 kg be stretched for transverse waves of frequency 43.0 hz to hav

e a wavelength of 0.800 m ?
Physics
1 answer:
Blababa [14]3 years ago
5 0

Solution:

v=sqrt(\frac{t}{u})\\v=40\times0.80=32m/s\\\frac{t}{u}=900\\t=900\times\frac{0.150}{2.20}\\=61.36N


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Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half-way between the two plates th
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Answer:

Explanation:

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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
ArbitrLikvidat [17]

Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

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From the question we are told that

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   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

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

      d_2  =  2 * d_1

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         A_1 v_1  =  A_2 v_2

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             A \ \  \alpha \ \  d^2

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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

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         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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