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Calculating g on Moon

An astronaut on the moon wishes to calculate the local g in the area. He sets up a wire with a mass of 1.7 grams and hanging from it an object with mass 4.98 kg. He sends some kind of pulse down the wire and calculates that it takes 40.86 ms to traverse the wire. What is g? You can neglect the mass of the wire when calculating the tension.

That was everything that was given in the problem. I have not a clue how to solve it. What I did was just use the equation of T=2pi[sqrt(l/g)] to solve for g and hope that the wire acts like a spring or something.

I’m totally sure that’s incorrect, but hey I’m not an astrophysicist.

 

(Response from Members and visitors awaited)

 

Posted by Ashish Dharva

Change in resistance of a wire on stretching

A WIRE OF LENGTH L AND AREA OF CROSS SECTION A HAS A RESISTANCE R. IF THE WIRE IS STRACHED TO TWICE OF ITS LENGTH WHILE A REMAINS SAME , WHAT IS THE NEW RESISTANCE????

Answer:

We know that R = ρl/A

When the wire is stretched, there is no change in volume. So if length is doubled then volume will be halved so that volume is constant.

Therefore, the new resistance R’ = 4R

That is the resistance becomes four times its original value.

1. A WIRE OF LENGTH L AND AREA OF CROSS SECTION A HAS A RESISTANCE R. IF THE WIRE IS STRACHED TO TWICE OF ITS LENGTH WHILE A REMAINS SAME , WHAT IS THE NEW RESISTANCE????

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Schrodinger’s Cat in Daily Life

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