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6.67 ptsQuestion 13What is the length of a pendulum that has - period of 1.500 s?33.34 cm12.82 cm55.85 cm6,210 cmQuestion 146.67 ptsswimming pool propagate at 0.75 ...

Question

6.67 ptsQuestion 13What is the length of a pendulum that has - period of 1.500 s?33.34 cm12.82 cm55.85 cm6,210 cmQuestion 146.67 ptsswimming pool propagate at 0.75 m/s: You splash the water at one end of the pool and Waves an observe the wave g0 to the opposite end, reflect ad return in 40.00 $. How far away is the other end of the pool?47,43 m15.C0m11.25m18.05 m

6.67 pts Question 13 What is the length of a pendulum that has - period of 1.500 s? 33.34 cm 12.82 cm 55.85 cm 6,210 cm Question 14 6.67 pts swimming pool propagate at 0.75 m/s: You splash the water at one end of the pool and Waves an observe the wave g0 to the opposite end, reflect ad return in 40.00 $. How far away is the other end of the pool? 47,43 m 15.C0m 11.25m 18.05 m



Answers

Waves on a swimming pool propagate at $0.75 \mathrm{m} / \mathrm{s}$. You splash the water at one end of the pool and observe the wave go to the opposite end, reflect, and return in 30.00 s. How far away is the other end of the pool?

So here we need to use the parallel access the room first, and we can say that the total moment of inertia would be equal to the moment of inertia through the center of mass. In this case, this would be 1/2 M r squared plus an H squared were here. H is gonna be equal to l plus R, and we know that our is equaling 0.15 meters. So we can rewrite this and say that this will be equal to 1/2 and r squared plus and multiplied by l plus our quantity squared. We know the mass here is gonna vehicle to 1.0 kilograms. And so, with the period equaling 2.0 seconds, we can then use equation 15 29 and say that here 2.0 seconds would be equal to two pi multiplied by of the moment of inertia 1/2 m r squared plus m times out, our quantity squared and this would be all divided by and g times h or again l plus are now we could all directly manipulate it will be a bit messy, but was much easier. Would simply be too. Um use your sole function in your T I t for 85 or 89 then substitute in 1.0 kilograms for m and then substituting in 0.15 meters for our and the period is already substituted for you 2.0 seconds. So essentially, l after using the salt function and solving for El Al is gonna vehicle 2.8315 meters. This would be our final answer. That is the end of the solution. Thank you for

So here the speed is the distance is, of course, the speed times time. And so here we know that we want to. The wave is essentially going to one end of the pool and then coming back to the opposite end. And so we can simply say that two d two times the distance would be equaling the speed, times the time. And of course, D would be equaling the length of the pool. And so we can say that then the distance or the length of the pool would be 0.750 meters per second. Most applied by the round trip time. 30.0 seconds divided by two. And so we find that the length of the pool is approximately 11.25 meters. This would be our final answer. That is the end of the solution. Thank you for watching

In this problem on the topic of oscillators, we're told that a seconds pendulum is a pendulum that can measure A period of exactly two seconds to each. One way swing is exactly 1/2. We want to calculate the length of this pendulum. In austin texas where we are given the acceleration due to gravity. We're then told the pendulum is moved to paris with the acceleration due to gravity is slightly different. And we want to know how how much by how much we need to lengthen this pendulum. And lastly we want to calculate the length of the second pendulum on the moon where again we are given the acceleration due to gravity. You know in general that the period of a pendulum T. Is linked to its length. By the equation T. Is equal to two pi times the square root of L. Over G. Where L. Is the length of the pendulum, which means the length of the pendulum, L. Is equal to T squared times G over four hi squared. So if we first take this equation to austin, the length of the pendulum in Austin, L. A. Is equal to T squared times the acceleration due to gravity at Austin G. A. Over four pi squared. So this is a period of exactly two seconds squared Times the value for G. Austin, which is 9.793 meters per square second derided by for hi squid. And this gives us the length of the pendulum will need to be a zero point 99 22 the eight m. Or keeping keeping our four significant figures zero point 99 22 meters. Now, if we move this pendulum to paris, we need to first calculate the length of this pendulum in paris LP and L. P. Is equal to T squared times the gravitational acceleration in paris G. P over four pi squared. And so if we substitute our values in again, the period Must remain at two seconds and that's squared. I'm the gravitational acceleration which is given to be 9.809 meters per square second, divided by for times pi squared. And so the length that the pendulum needs to be in paris is zero point 99 38 59 m again keeping Or four significant figures zero point 99 three nine m. And so the learned that the pendulum needs to be increased. Delta L. Is the length that's needed in paris minus tellin needed in Austin, which is zero point 99 3859 minus zero point 99 22 the eight. This gives us The increased length in m to be zero 00 16 21 m or one 0.6 millimeters. So we need to increase the length of the pendulum By 1.6 mm when going from Austin to Paris, in order to keep the period at two seconds. Lastly for the moon, the length of this pendulum, our M is equal to the period squared t squared times the gravitational acceleration of the moon GM Over four Pi squared. So again, this is A period of two seconds squared Times a gravitational acceleration on the moon of 1.62 meters per square second, divided by four times pi squared. And so calculating, we get the length of this pendulum on the moon To be 0.164 m.

The period of a pendulum is given by T is equal to two pi and the square root of allergy, and hence the length of opinion of From here, everybody doing prearranged equation is T Square MSG. But we're full pie squid and so fusty we can calculate the limp off pendulum in Austin who call it Al A On this limb is equal to periods were multiplied right by the gravitational acceleration in Austin G eight divided by a full pi squared. These values are known so European pendulum is two seconds. Mr Pressed the humans. Yeah, that's quit. Times got is an acceleration in Austin is measured to be 9.7 nine d You just square sinking all over for Pike Squid. And this gives us Linda the pendulum in Austin to be zero point right nine to to, I mean just next we want to calculate 1,000,000,000 off the pendulum in Paris, so call it called P. And now Pete to again the P it's were once black of the gravitation acceleration Paris GP over for high script and again, these values and no. So that's Pierre off two seconds. Well squared. What it by by the given Gravitation Exploration Paris 9.8 09 meters per square second all over for I squared and being calculation. This gives us the Linden Paris to be zero point 99 No meet us And so the difference between these two billions required in Paris minus the length of the pendulum in Austin P minus out a is equal to 0.99 39 meters minus in Austin off Benjamin, 0.9 to two meters. This gives us shortening in Austin off 1.6 really means So when you from Paris to Austin Bill into the pendulum must be shortened my 1.6 millimeters in order to maintain it to second period Finally, Ngilu from the moon call it out. You can use the same creation for that's t script NZ gravitation exploration on the moon you might buy full price were this is equal to again impeded two seconds squared rotational acceleration of who is 1.62 You just square second This is all over for fight square. During the calculation, you see the length off The pendulum on the moon is equal to zero 0.1 six. Hold meet


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