a 80 gram tennis ball is attached to a 1.1 meter string of negligible mass and swung in a 0.7 meter radius horizontal circle. Find the magnitude of the tension force. 

To solve for the magnitude of the tension force in the string, we'll apply concepts from circular motion. Here’s the approach:

### 1. Identify Given Information:
- **Mass of the tennis ball**: \( m = 80 \, \text{g} = 0.080 \, \text{kg} \)
- **Length of the string**: \( L = 1.1 \, \text{m} \)
- **Radius of the circular path**: \( r = 0.7 \, \text{m} \)

Since the ball moves in a horizontal circle, the tension in the string provides the centripetal force required for circular motion. This force can be calculated using the following steps.

### 2. Break Down the Problem:
The string is inclined at an angle because the radius of the circle (\( r = 0.7 \, \text{m} \)) is less than the full length of the string (\( L = 1.1 \, \text{m} \)). Let’s denote the angle between the string and the vertical as \( \theta \).

We can use trigonometry to find the angle \( \theta \):
\[
\cos(\theta) = \frac{r}{L} = \frac{0.7}{1.1} \approx 0.636
\]
Thus,
\[
\theta = \cos^{-1}(0.636) \approx 50.41^\circ
\]

### 3. Equilibrium of Forces:
For the ball to stay in circular motion, the tension force \( T \) in the string has two components:
- The **vertical component** balances the gravitational force:
  \[
  T \cos(\theta) = mg
  \]
  where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
- The **horizontal component** provides the centripetal force:
  \[
  T \sin(\theta) = \frac{mv^2}{r}
  \]
  where \( v \) is the velocity of the ball.

### 4. Solve for \( T \) (Tension Force):
First, solve for \( T \) from the vertical component equation:
\[
T = \frac{mg}{\cos(\theta)}
\]
Substitute the values:
\[
T = \frac{0.080 \times 9.8}{\cos(50.41^\circ)} \approx \frac{0.784}{0.636} \approx 1.232 \, \text{N}
\]

Thus, the magnitude of the tension force is approximately **1.23 N**.

### Conclusion:
The tension in the string is \( T \approx 1.23 \, \text{N} \).

Would you like further details on this solution or clarifications? 

Here are five related questions:
1. How do you calculate the velocity of an object in circular motion?
2. What role does centripetal force play in horizontal circular motion?
3. How does changing the radius affect the tension in the string?
4. Why is the string's mass considered negligible in this scenario?
5. Can you explain the relationship between tension and the angle of the string?

**Tip**: Always use consistent units (SI units) to avoid conversion mistakes during calculations.

A 80 gram tennis ball is attached to a 1.1 meter string of negligible mass and swung in a 0.7 meter radius horizontal circle. Find the magnitude of the tension force.

Learn how to calculate the tension force of a tennis ball attached to a string and swung in a horizontal circular motion. The solution applies concepts of circular motion, trigonometry, and forces, including tension and centripetal force. Ideal for students in Grades 11-12 studying advanced physics.

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