3. Effect of Varying Mass 2 Set Mass 1 to be 10 kg, the location of Mass 1 to be 0 meters, and the location of Mass 2 to be 5 meters. Complete the following table. As an example, the third column, "Ratio of force" is computed using the formula: Ratio of force = (Force when Mass 2 is 10 kg)/(Force when Mass 2 is 20 kg) %3D when Mass 2 is 20 kg. Mass 2 Force Ratio of force 10 kg 20 kg 40 kg 100 kg 267x10 5.34r1070 1.07x 10- 2.67x10-LO 1.0 0.5 2-5 What do you notice about the "Ratio of forces" when the mass of Mass 2 Doubles? at 406g the mass Doubles and at 100 14g the mass Triples? No 90 is the Same. Increases by four times? N What do expect the "Ratio of forces" to be when the Mass 2 is increased by 50 times?

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### Effect of Varying Mass 2

Set Mass 1 to be 10 kg, the location of Mass 1 to be 0 meters, and the location of Mass 2 to be 5 meters. Complete the following table.

As an example, the third column, "Ratio of force," is computed using the formula:

\[ \text{Ratio of force} = \frac{\text{Force when Mass 2 is 10 kg}}{\text{Force when Mass 2 is 20 kg}} \]

when Mass 2 is 20 kg.

| Mass 2 | Force          | Ratio of force |
|--------|----------------|----------------|
| 10 kg  | \(2.67 \times 10^{-10}\) | 1.0            |
| 20 kg  | \(5.34 \times 10^{-10}\) | 0.5            |
| 40 kg  | \(1.07 \times 10^{-9}\)  | 0.25           |
| 100 kg | \(2.67 \times 10^{-9}\)  | 0.1            |

#### Observations:
- Question: What do you notice about the "Ratio of forces" when the mass of Mass 2 doubles?
  - **Answer:** At 40 kg, the mass doubles, and the ratio is halved. At 100 kg, it follows the pattern, maintaining consistency with doubling and halving.

- Question: What do you expect the "Ratio of forces" to be when the Mass 2 is increased by 50 times?
  - **Answer:** This part seems unanswered in the image, but following the pattern, it would continue to decrease proportionally as the mass increases.
Transcribed Image Text:### Effect of Varying Mass 2 Set Mass 1 to be 10 kg, the location of Mass 1 to be 0 meters, and the location of Mass 2 to be 5 meters. Complete the following table. As an example, the third column, "Ratio of force," is computed using the formula: \[ \text{Ratio of force} = \frac{\text{Force when Mass 2 is 10 kg}}{\text{Force when Mass 2 is 20 kg}} \] when Mass 2 is 20 kg. | Mass 2 | Force | Ratio of force | |--------|----------------|----------------| | 10 kg | \(2.67 \times 10^{-10}\) | 1.0 | | 20 kg | \(5.34 \times 10^{-10}\) | 0.5 | | 40 kg | \(1.07 \times 10^{-9}\) | 0.25 | | 100 kg | \(2.67 \times 10^{-9}\) | 0.1 | #### Observations: - Question: What do you notice about the "Ratio of forces" when the mass of Mass 2 doubles? - **Answer:** At 40 kg, the mass doubles, and the ratio is halved. At 100 kg, it follows the pattern, maintaining consistency with doubling and halving. - Question: What do you expect the "Ratio of forces" to be when the Mass 2 is increased by 50 times? - **Answer:** This part seems unanswered in the image, but following the pattern, it would continue to decrease proportionally as the mass increases.
Expert Solution
Step 1

Given:

The mass m1=10kg.

The position of mass m1 from origin is 0 m.

The initial mass m2=10kg.

The position of mass m2 from origin is 5 m.

Thus, the distance between m1 and m2 is d=5 m.

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