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An astronaut on the Moon wishes to measure the local value of $g$ by timing pulses traveling down a wire that has a large object suspended from it. Assume a wire of mass 4.00 $\mathrm{g}$ is 1.60 $\mathrm{m}$ long and has a $3.00-\mathrm{kg}$ object suspended from it. A pulse requires 36.1 $\mathrm{ms}$ to traverse the length of the wire. Calculate $g_{\text { Moon }}$ from these data. (You may neglect the mass of the wire when calculating the tension in it.)

1.64$\mathrm { m } / \mathrm { s } ^ { 2 }$

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linear density for the wire would be 4.0 times 10 to the negative third kilograms. This would be divided by 1.6 meters and this is giving us 2.5 times 10 to the negative third kilograms per meter. We have the speed of the Wave V being 1.6 meters in 36.1 times 10 to the negative third seconds. And so this is giving us 44 0.3 meters per second. And we can say that then, um, the velocity, the speed of the transverse way would be equal to the square root of the force under tension s divided by the linear density mu and in this case, the forces, the gravitational force. So this would be the mass multiplied by the gravity on the moon. This would be divided by the Lanier density mu. And so we can say that Then the gravity on the moon is equaling the linear density times a velocity squared divided by the mass M. And so we can say that then gravity on the moon will be equal to 2.5 times 10 to the negative third kilograms per meter multiplied by 44.3 meters per second. Quantity squared, divided by 3.0 kilograms. And we find that then the gravity on the moon is one 0.64 meters per second squared. This is our final answer. That is the end of the solution. Thank you for watching.

Carnegie Mellon University