away and has a mass of 60. kg. 10 Find the resultant force on (a) the 0.100 kg mass and (b) the 0.200 kg mass in Fig. 6-15 (the masses are iso- lated from the earth). 0.400 kg 0.200 kg 0.100 kg 10.0 cm 10.0 cm Fig. 6-15

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### Physics Problem: Gravitational Force Calculation

#### Task Description:
**Problem 10:** Find the resultant force on:
- (a) the 0.100 kg mass
- (b) the 0.200 kg mass in Fig. 6-15 (the masses are isolated from the earth).

#### Diagram Explanation:
**Figure 6-15:** This figure displays three masses aligned in a straight line and isolated from any external gravitational influences, including Earth.

- The mass on the left is **0.400 kg**.
- The mass in the center is **0.200 kg**.
- The mass on the right is **0.100 kg**.

The distances between the masses are as follows:
- The distance between the 0.400 kg and the 0.200 kg masses is **10.0 cm**.
- The distance between the 0.200 kg and the 0.100 kg masses is also **10.0 cm**.

### Further Problem:
**Problem 11:** Three isolated particles, each having a mass of 2.00 kg, are at the vertices of an equilateral triangle with 1.00 m sides. Find the magnitude and direction of...

(NOTE: The rest of Problem 11 is cut off in the image.)

### Educational Context:
This problem requires students to apply their knowledge of gravitational forces between masses, including the use of Newton's Law of Universal Gravitation, which states:

\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

where:
- \(F\) is the gravitational force,
- \(G\) is the gravitational constant (\(6.67430 \times 10^{-11} \ \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}\)),
- \(m_1\) and \(m_2\) are the masses of the two objects,
- \(r\) is the distance between the centers of the two masses.

### Solution Strategy:
1. **Calculate the gravitational force** between the 0.400 kg and 0.200 kg masses and the 0.200 kg and 0.100 kg masses.
2. **Consider the directions** of these forces to determine the resultant force on each mass.
3. **Account for the Vector Addition** of forces when summing the gravitational influences on each specific mass.

Understanding
Transcribed Image Text:### Physics Problem: Gravitational Force Calculation #### Task Description: **Problem 10:** Find the resultant force on: - (a) the 0.100 kg mass - (b) the 0.200 kg mass in Fig. 6-15 (the masses are isolated from the earth). #### Diagram Explanation: **Figure 6-15:** This figure displays three masses aligned in a straight line and isolated from any external gravitational influences, including Earth. - The mass on the left is **0.400 kg**. - The mass in the center is **0.200 kg**. - The mass on the right is **0.100 kg**. The distances between the masses are as follows: - The distance between the 0.400 kg and the 0.200 kg masses is **10.0 cm**. - The distance between the 0.200 kg and the 0.100 kg masses is also **10.0 cm**. ### Further Problem: **Problem 11:** Three isolated particles, each having a mass of 2.00 kg, are at the vertices of an equilateral triangle with 1.00 m sides. Find the magnitude and direction of... (NOTE: The rest of Problem 11 is cut off in the image.) ### Educational Context: This problem requires students to apply their knowledge of gravitational forces between masses, including the use of Newton's Law of Universal Gravitation, which states: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] where: - \(F\) is the gravitational force, - \(G\) is the gravitational constant (\(6.67430 \times 10^{-11} \ \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}\)), - \(m_1\) and \(m_2\) are the masses of the two objects, - \(r\) is the distance between the centers of the two masses. ### Solution Strategy: 1. **Calculate the gravitational force** between the 0.400 kg and 0.200 kg masses and the 0.200 kg and 0.100 kg masses. 2. **Consider the directions** of these forces to determine the resultant force on each mass. 3. **Account for the Vector Addition** of forces when summing the gravitational influences on each specific mass. Understanding
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