Asked by Ray Scholz
You're on planet YZ and an astronaut observes that a 1.0 m long pendulum has a period of 1.50s. What is the free-fall accerlerttion on YZ?
Answers:
1.003 m/s^2
26.4 m/s^2
17.6 m/s^2
0.23 m/s^2
Could someone direct me how to set this up or give me any assistance, please?
Answers:
1.003 m/s^2
26.4 m/s^2
17.6 m/s^2
0.23 m/s^2
Could someone direct me how to set this up or give me any assistance, please?
Answers
Answered by
Henry
T^2 = 4*(3.142)^2*(L/g) = 2.25
39.48(1/g) = 2.25,
39.48/g = 2.25,
g = 39.48 / 2.25 = 17.6m/s^2.
39.48(1/g) = 2.25,
39.48/g = 2.25,
g = 39.48 / 2.25 = 17.6m/s^2.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.