To determine the period of a pendulum on Mars, we use the formula for the period [tex]\( T \)[/tex], which is given by:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- [tex]\( T \)[/tex] is the period of the pendulum,
- [tex]\( \pi \)[/tex] (pi) is approximately 3.14159,
- [tex]\( L \)[/tex] is the length of the pendulum (1.8 meters in this case),
- [tex]\( g \)[/tex] is the acceleration due to gravity (3.69 meters/second[tex]\(^2\)[/tex] on Mars).
Substitute [tex]\( L \)[/tex] and [tex]\( g \)[/tex] into the formula:
[tex]\[ T = 2\pi \sqrt{\frac{1.8}{3.69}} \][/tex]
Next, perform the calculations in steps:
1. Calculate the division inside the square root:
[tex]\[ \frac{1.8}{3.69} \][/tex]
2. Take the square root of the result from step 1:
[tex]\[ \sqrt{\frac{1.8}{3.69}} \][/tex]
3. Multiply the result by [tex]\( 2\pi \)[/tex]:
[tex]\[ 2\pi \sqrt{\frac{1.8}{3.69}} \][/tex]
The solution to these calculations gives us the period of the pendulum. The correct period is approximately 4.4 seconds.
Thus, the correct answer is:
O E. 4.4 seconds
