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A pendulum on Earth has a period of 5.70 s. What would theperiod be of that same pendulum on mythical planet X that has agravitational acceleration of 2.33 m/s2? Ex...

Question

A pendulum on Earth has a period of 5.70 s. What would theperiod be of that same pendulum on mythical planet X that has agravitational acceleration of 2.33 m/s2? Express your answer inseconds.

A pendulum on Earth has a period of 5.70 s. What would the period be of that same pendulum on mythical planet X that has a gravitational acceleration of 2.33 m/s2? Express your answer in seconds.



Answers

A pendulum has a period of 1.85 s on Earth. What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth?

Properly in 39. Uh, we here we have here a relation between the tea that is directly proportional with the square root of the length and university proportion with the square root of the acceleration. So we need first to can create the K in order to complete any of the T values. So the key, which is the constant, can be calculated using the givens. So we know that he is equal to six. Uh, when the length is equal to 289 centimeters and the G or the acceleration is equal to 900 80. So the key is equal to, um so the K is equal to 34.711 So the second point we need to calculate the period t um, using the same constant when the length is equal to 121 and the acceleration is equal to 980. So this is equal to 1 to 197 seconds

All right, take to, um period of a pendulum will call it p very directly with square root of the length of the pendulum on inversely with the square root of the acceleration due to gravity. Ok, Um, well, the acceleration due to gravity at Not very well I mean, that's that's that's ah, constant. So that should make things a little bit easier. Um, just find the period when the length is 1 21 The acceleration due to gravity is 9 80 uh, 70 for square foot. If the period is six pie, when the, um it's the length is to 89 the acceleration due to gravity is 9 18 of that 9 80 is the same, um, both sides. So I don't try Make this easy and I think easier might be not like simplifying all the way. So So we'll do. Square 2 to 89 is 17 so 17 k over the square root of 9 80 And I think I believe that's not a perfect square. So just leave it, um, so if I multiply both sides by the square root of 9 80 over 17 and get seven teams to cancel and get this squared of nine eighties to cancel. So the square root of 9 80 over 17. So let's call it six pie. Let's use my black here again. So six pi squared 9 80 over 17. That's r K value. Don't worry. I think it's gonna or got. Okay, you're in and all right, so we're supposed to find out the period, Um, if the So you have six pie sward of 9 80/17 if the length is a squirt of 1 21 and the acceleration due to gravity is the squared of 9 80 again. All right, so let's simplify. Um, squirt of 1 21 is 11. 11 times six is 66. So 66 pi times a squirt of 9 80 all over 17 all over the squared of 9 80 If I oh, get rid fractions. Um, I could, you know, multiply this by 17 and multiply this by 17. I would Mac those out discount blacks for action work there with me. Um, then we'd have 60 six high times the square root of 9 80 all over the squared of 9 80 times 17 and then the squared of nine eighties. We could see that those who cancel out and we finally there's probably a better way this works, guys, we get the 66 pi over 17. Um, what's that measured in seconds? So that's the period.

Repeated of a pendulum is given by t is equal to pi squared L over G and the land is assumed to be the same for the pendulum board on Mars and so we can find the ratio off the period of the pendulum aun mas t m over the period when up teeth and this is equal to two pi times square root hell the length of the pendulum Divide it by the gravitation examination when mods g m all divided by appear of opinion among us which is to pine and the square root lentil depend You help over jean the gravitational acceleration and so we can see that things can sell off and we get that this ratio is the square root off the gravitation acceleration on, uh divided by the gravitational acceleration one month and therefore we can see at the PD A of motion on mas is equal to the puke off the motion of the oscillation and and the square root off the ratio off the gravitation Accelerations g e oh, and she him These varies all known so we can calculate what the pier and beyond months So here up 0.8 sanctions. Times Square root the temptation exploration credits, accreditation exhortations. It was 1/0 0.37 The gravitational acceleration on Mars is 37% and hence we get repeated off the estimation on mothers to be one point the seconds.

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.


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