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After landing on an unfamiliar planet, a space explorerconstructs a simple pendulum of length 47.0 cm c m . The explorerfinds that the pendulum completes 104 full s...

Question

After landing on an unfamiliar planet, a space explorerconstructs a simple pendulum of length 47.0 cm c m . The explorerfinds that the pendulum completes 104 full swing cycles in a timeof 126 s s .What is the magnitude of the gravitational acceleration on thisplanet?Express your answer in meters per second per second.

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 47.0 cm c m . The explorer finds that the pendulum completes 104 full swing cycles in a time of 126 s s . What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per second.



Answers

Astronauts on a distant planet set up a simple pendulum of length 1.2 $\mathrm{m}$ . The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

As we all know that the time period is given by the formula is equal to two pi route under L by G planet. So solving it for them, I can write the value of the ad G. Planet is equal to poor pie is square multiplication and L by the square. So just putting the value here, I can only express an age, four pi squared multiplication and 2.20 m by 2.87 2nd. Holy Square Return simplification. I get devaluate 10 points 54 m per second, the square as the gravitational acceleration on the planet. So I hope you understand how I solve this problem with the help of simple concept and general formula. So this is our final and said 10.54 m/s squared.

We know the time period T. Of a simple pendulum with equal to two pi, multiplied by the square root of the length, divided by the acceleration due to gravity, solving for the acceleration due to gravity. This would be equaling 24 pi squared L. Divided by time period squared. So G would be equaling 24 pi squared, multiplied by the length of the curriculum At 0.5 m .50 m. This would be divided by 1.50 seconds quantity squared, and the acceleration on the planet rounded to two significant figures 8.8 m per second squared acceleration to the gravity on the unknown planet. That is the end of the solution. Thank you for watching.

All right. So the question was asking us what disappearance off the pendulum on surface off the Mars. Okay, so we know appear on the surface of Mars. TM is equal to two pi times square throughout. Oh, GM, our zalando pendulum gm is the gravitational constant on Mars. Well, we know the gravitational constant on Mars, which is 3.71 meter per second squared. Okay, we wasn't on the time. Here on Earth is 1 26 06 So it seems I we need to determine appearance on Mars, willing to determine the lands of pension, and we can determine the lens opinion on the time here equation on earth, which is t equals two pi times square rules. Oh, gee and G, here's a gravitational constant on earth. Okay, so you reduce our arrangement we get at El is just simply go to gee tens t e square. Oh, we're who Hi. Oh, by square. Okay. And this will give us g member was not going a meter per second squared trance, Theo, time here on earth, which is 1.60 seconds square. Okay, so over full Baesler. And this will give us the land off. The pendulum is about zero boy, 64 meter. What? So now if we plug him back totally time here on Mars equation will get TM is equal to to buy Times Square. Out was throw going 64 meaner over the gravitational constant on Mars, which is 3.71 meter per second square. And this will give us the time period on Mars is equal to and me was 2.6 second. Okay, so unless the answer for this question Thank you.

The period of a prison is given by He is equal to two pi a route under al by key solving it further, I cannot devalue G is equal to and multiplication. Bye bye. T. Foley Square. So I will just put the value in this X. Person so I can write, the value of G is equal to 0.50 m multiplication to pay by 1.5 years again. Holy Square which is equal to 8.8 meter per second. Holy Square. As the final answer for this problem. I hope you understand the solution. I'm reporting the concept once again. First I just tried the formula. After that I just changed the rearrange the equation to get the value of Z. And finally I just put all the value. I just put the value of L two pi and th 1.50 per second 2nd. And after that finally I get this value as the answer.


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