Answer :
To find the period of a pendulum on Mars where the gravity is 3.69 meters/second² and the length of the pendulum is 1.8 meters, we would use the formula for the period of a simple pendulum:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
1. Identify the given parameters:
- [tex]\( g \)[/tex] (gravity on Mars) = 3.69 meters/second²
- [tex]\( L \)[/tex] (length of the pendulum) = 1.8 meters
2. Substitute the values into the formula:
[tex]\[ T = 2\pi \sqrt{\frac{1.8}{3.69}} \][/tex]
3. Calculate the value inside the square root:
[tex]\[ \frac{1.8}{3.69} \approx 0.4878 \][/tex]
4. Take the square root of the result:
[tex]\[ \sqrt{0.4878} \approx 0.6984 \][/tex]
5. Multiply by [tex]\( 2\pi \)[/tex]:
[tex]\[ T \approx 2 \times \pi \times 0.6984 \approx 4.388 \text{ seconds} \][/tex]
Thus, the period of the pendulum on Mars would be approximately 4.4 seconds.
The correct answer is:
E. 4.4 seconds
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
1. Identify the given parameters:
- [tex]\( g \)[/tex] (gravity on Mars) = 3.69 meters/second²
- [tex]\( L \)[/tex] (length of the pendulum) = 1.8 meters
2. Substitute the values into the formula:
[tex]\[ T = 2\pi \sqrt{\frac{1.8}{3.69}} \][/tex]
3. Calculate the value inside the square root:
[tex]\[ \frac{1.8}{3.69} \approx 0.4878 \][/tex]
4. Take the square root of the result:
[tex]\[ \sqrt{0.4878} \approx 0.6984 \][/tex]
5. Multiply by [tex]\( 2\pi \)[/tex]:
[tex]\[ T \approx 2 \times \pi \times 0.6984 \approx 4.388 \text{ seconds} \][/tex]
Thus, the period of the pendulum on Mars would be approximately 4.4 seconds.
The correct answer is:
E. 4.4 seconds
