Answer :
To determine the centripetal acceleration of a small laboratory centrifuge where the tip of the test tube is moving at 19.0 meters per second in a circle with a radius of 10.0 centimeters, we can follow these steps:
1. Understand the given values:
- Velocity ([tex]\(v\)[/tex]): 19.0 meters/second
- Radius ([tex]\(r\)[/tex]): 10.0 centimeters, which needs to be converted to meters. Since 1 meter = 100 centimeters, we have [tex]\(r = 0.1\)[/tex] meters.
2. Recall the formula for centripetal acceleration ([tex]\(a\)[/tex]):
The centripetal acceleration is given by the equation:
[tex]\[ a = \frac{v^2}{r} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ a = \frac{(19.0 \ \text{meters/second})^2}{0.1 \ \text{meters}} \][/tex]
4. Calculate the acceleration:
- First, square the velocity: [tex]\(19.0 \times 19.0 = 361.0 \ \text{meters}^2/\text{second}^2\)[/tex].
- Next, divide by the radius: [tex]\(\frac{361.0 \ \text{meters}^2/\text{second}^2}{0.1 \ \text{meters}} = 3610.0 \ \text{meters/second}^2\)[/tex].
The centripetal acceleration is [tex]\(3610.0 \ \text{meters/second}^2\)[/tex].
Given the multiple-choice options:
- Option A: 1.82 x 10 meters/second didn't match.
- Option B: 3.61 x 10 meters/second looks close, as [tex]\(3.61 \times 10^3\)[/tex] meters/second² matches our result numerically (3610.0 meters/second²).
- Option C: 5.64 x 10 meters/second didn't match.
- Option D: 2.49 x 10 meters/second didn't match.
- Option E: 1.18 x 10 meters didn't match.
The correct answer is:
Option B: 3.61 x 10 meters/second.
1. Understand the given values:
- Velocity ([tex]\(v\)[/tex]): 19.0 meters/second
- Radius ([tex]\(r\)[/tex]): 10.0 centimeters, which needs to be converted to meters. Since 1 meter = 100 centimeters, we have [tex]\(r = 0.1\)[/tex] meters.
2. Recall the formula for centripetal acceleration ([tex]\(a\)[/tex]):
The centripetal acceleration is given by the equation:
[tex]\[ a = \frac{v^2}{r} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ a = \frac{(19.0 \ \text{meters/second})^2}{0.1 \ \text{meters}} \][/tex]
4. Calculate the acceleration:
- First, square the velocity: [tex]\(19.0 \times 19.0 = 361.0 \ \text{meters}^2/\text{second}^2\)[/tex].
- Next, divide by the radius: [tex]\(\frac{361.0 \ \text{meters}^2/\text{second}^2}{0.1 \ \text{meters}} = 3610.0 \ \text{meters/second}^2\)[/tex].
The centripetal acceleration is [tex]\(3610.0 \ \text{meters/second}^2\)[/tex].
Given the multiple-choice options:
- Option A: 1.82 x 10 meters/second didn't match.
- Option B: 3.61 x 10 meters/second looks close, as [tex]\(3.61 \times 10^3\)[/tex] meters/second² matches our result numerically (3610.0 meters/second²).
- Option C: 5.64 x 10 meters/second didn't match.
- Option D: 2.49 x 10 meters/second didn't match.
- Option E: 1.18 x 10 meters didn't match.
The correct answer is:
Option B: 3.61 x 10 meters/second.