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tigry1 [53]
3 years ago
10

A 20-cm long spring is attached to the wall. When pulled horizontally with a force of 100N, the spring stretches to a length of

22cm. What is the value of the constant? A.) The same spring is used in a tug of war. Two people pull on the ends, each with a force of 100N. How long is the stretched string? B.) The same spring is now suspended from a hook and a 10.2kg block is attached to the bottom end. How long is the stretched spring?
Physics
1 answer:
suter [353]3 years ago
3 0

Answer:

Explanation:

Part 0

All the spring moves is 2 cm

x = 2 cm * [1 m / 100 cm ]

x = 0.020 meters

F = k*d

100N = k * 0.02 m

100 N / 0.02 = k

5000 N / m

Part A

The spring feels a force of 100 N - - 100N = 200 N because each person is pulling in the opposite direction.

F = k * x

200N = 5000 N/m * d

200 / 5000 = d

d = 0.04 meters.

Part B

10.2 kg must be converted to a force as experienced here on earth.

F = m * g

g = 9.81

m = 10.2

F = 10.2 * 9.81

F = 100.06 N

F = k * d

100.06 = 5000 * d

d = 100.06 / 5000

d = 0.02 meters.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

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

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

Explanation:

Mathematically Frequency can be represented as

                         F = \frac{v}{\lambda}

Where \lambda is the wavelength and v is the velocity

   Now looking at the diagram we see that

          For the  first frequency we have

             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}

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

  So,

        For the  first frequency we have

                                 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

                               v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F

Hence

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

 Mathematically angular speed can be represented as

                           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  

 Hence

          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  

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

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                    =>    r = \frac{v}{w}

 Since the linear velocity is constant we have that

                            r \  \alpha \  \frac{1}{w}

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

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