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lorasvet [3.4K]
2 years ago
15

Q15.17 Both wave intensity and gravitation obey inverse-square laws. Do they do so for the same reason?Discuss the reason for ea

ch of these inverse-square laws as well as you can. block is attached to an ideal spring and is mov ing in SHM on a horizontalfrictionless surfaceThe amplitude of the motion is 0.120 m.The maximum speed of the block is 3.90 m/s.What is the maximum magnitude of the acceleration of the block?
Physics
1 answer:
s2008m [1.1K]2 years ago
7 0

(a) Intensity obeys inverse square law from basis of light passing through a given surface.

(b) Gravitation obeys inverse square law from the basis of force between two masses.

(c) The maximum magnitude of the acceleration of the block is 126.75 m/s².

<h3>What is intensity?</h3>

Intensity is the ratio is ratio of power to area of a given surface.

I = P/A (W/m²)

where;

  • P is power
  • A is area
<h3>Universal gravitation law</h3>

F = \frac{Gm_1m_2}{r^2}

Intensity and gravitation do not obey inverse square law for same reason;

  • Intensity obeys inverse square law from basis of light passing through a given surface.
  • Gravitation obeys inverse square law from the basis of force between two masses.
<h3>Acceleration of the block</h3>

a = v²/A

a = (3.9²)/0.12

a = 126.75 m/s²

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

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The  z-component of the force is  \= F_z  =  0.00141 \ N    

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From the question we are told that

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           The  velocity of the particle toward the z-direction is

v_y  =  -5.85*10^{4}\ \r k  \ m/s

Generally the force on this particle is mathematically represented as

          \= F  =  q (\= v   X  \= B )

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          \= F  =  q ( v_x \r  i + v_y \r  j  +  v_z \r k  )  \ \ X \ (  \= B i)

         \= F  = q (v_y B(-\r  k) + v_z B\r j)      

  substituting values

       \= F  = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r  z) + [(5.58*10^{4}) (0.700)]\r y)    

      \= F=  0.00303\ \r j +0.00141\ \r k                  

So the z-component of the force is  \= F_z  =  0.00141 \ N    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).

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Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

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Explanation:

a) We have equation of motion v = u + at

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  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

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3 years ago
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