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dmitriy555 [2]
3 years ago
8

How long does it take a wave to travel 1200 meters with the speed of 3 X 108m/sec?

Physics
1 answer:
Vladimir79 [104]3 years ago
5 0
Around 3.70 seconds unless traveling through media

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Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used
natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is  T =8391.6 N

Explanation:

From the question we are told that

       The strain on the strain on the head is \Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m

         The contact area is  A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2  

Looking at the first diagram

           At  600 MPa of stress

               The strain is  0.3mm/mm

          At   450 MPa of stress

                 The strain is   0.0015 mm/mm

 To find the stress at  \Delta l we use the interpolation method

            \frac{\sigma_{\Delta l} -  \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l }  - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}

Substituting values

              \frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}

            \sigma _{\Delta l} -450 = 49.50

             \sigma _{\Delta l} = 499.50 MPa

Generally the force on each head is mathematically represented as

              F = \sigma_{\Delta l} * A

Substituting values

             F = 499.50*10^{6} * 2.8*10^{-6}

                =1398.6N

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

              T = 6 * F

                  = 6* 1398.6

              T =8391.6 N

                 

     

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3 years ago
As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t
Andreyy89

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

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You knew that this question is ridiculously easy.  So, just to
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It’s gonna be Oxygen ....
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Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.

The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

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