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Tcecarenko [31]
3 years ago
11

A mason stretches a string between two points 70 ft apart on the same level with a tension of 20 lb at each end. If the string w

eighs 0.18 lb, determine the sag h at the middle of the string.
Physics
1 answer:
sergejj [24]3 years ago
4 0

Answer:

Explanation:

Let tension be T in each string . Let angle with horizontal be θ in the middle

The sum of vertical components of tension of two string will balance the weight

2Tsinθ = mg

2 x 20 sinθ = .18

sinθ = .18 / 40 = .0045

θ = .25783 degree

If sag be y

y / 70 = tan .25783

y = 70 x tan.25783

= .3150 m

= 31.50 cm .

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The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

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Generally the distance of the listener to the first speaker is mathematically represented as

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Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

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         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

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