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valentinak56 [21]
3 years ago
6

A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5m long. The bowler hears the sound o

f the ball hitting the pins 2.50s after the ball is released from his hands. What is the speed of the ball? The speed of the sound is 340 m/s.
Physics
1 answer:
kobusy [5.1K]3 years ago
6 0
  <span> The ball takes 15/V secs to reach the pin since T=D/V. so there for the  Sound takes 15/340 secs to reach the bowler. The total time enlapsed then is 15/V + 15/340 = 2.10 So 15/V =2.1 -15/340
V = 7.296 m/s </span>
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Answer:

Please find the answer in the explanation

Explanation:

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3 years ago
The y-component of the force F which a person exerts on the handle of the box wrench is known to be 86 lb. Determine the x-compo
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Fy of the force F exerted on the handle of the box wrench = 86 lb

Considering the triangle in Fig 1

magnitude of perpendicular = P =  12

magnitude of base = B = 5

using Pythagoras theorem

                        H= \sqrt{P^{2} + B^{2}}

                 H= \sqrt{12^{2} + 5^{2}}

            H=13\\\implies cos \theta = \frac{5}{13} \\\implies sin \theta = \frac{12}{13}\\

y-component of force is given given as:

                              86 = Fsin\theta\\F = 86 (\frac{13}{12})\\F = 93.16 lb\\F_{x} =F cos \theta\\F_{x} = 93.16 (\frac{5}{13})\\F_{x} =38.18 lb.

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2 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

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 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
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