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schepotkina [342]
3 years ago
10

A heavy boy and a lightweight girl are balanced on a massless seesaw. The boy moves backward, increasing his distance from the p

ivot point. What happens to the seesaw? It is impossible to predict without knowing additional information. The side the boy is sitting on will tilt downward. The side the girl is sitting on will tilt downward. Nothing; the seesaw will remain balanced.
Physics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

The side the boy is sitting on will tilt downward.

Explanation:

According to the law of moments when the same force is applied at a greater distance from the pivot then the effect of moment is greater about that point.

<u>Mathematically momentum is given as:</u>

M=F\times r

where:

F is the applied force at a distance 'r' acting in a direction perpendicular to the line joining the point of application and the hinge.

  • Moment is the rotational effect of the applied force on the body.

<em>When the boy of a heavier mass than the girl was sitting on a balanced see-saw then it is certain that he was closer to the hinge than the girl to balance the turning effect (in case of an unbiased see-saw). When the body moves farther his weight is same but the radial distance from the hinge increases which increases his moment of weight.</em>

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

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b

 The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

                                w_1 >w_2 = w_3  

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The ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

                          r_2 =r_3 >r_1

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             Let the wavelength be  \lambda_1 = \lambda , and the frequency  F_1 = F

           For  the second frequency

           Let the wavelength be  \lambda_2 = 2 \lambda , and the frequency F_2 = \frac{F}{2}

           For  the third frequency

           Let the wavelength be  \lambda_3 = 2\lambda ,  and the frequency F_3 = \frac{F}{2}

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                                 v_1 = \lambda_1 F_1 = \lambda F

          For  the second frequency

                               v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F      

           For  the third frequency

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The Ranking of the curve according to their speed would be equal Rank because    v_1 =v_2 =v_3

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                           w = 2 \pi f

   For the  first frequency we have

                          w_1 = 2\pi F_1 = 2 \pi F                        

    For  the second frequency

                        w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2}  = \pi F

     For  the third frequency

                      w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2}  = \pi F  

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                          r_2 =r_3 >r_1

       

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