Accuracy of Gravity: -9.8m/s/s +- _______m/s/s

  • Thread starter Thread starter StephenDoty
  • Start date Start date
  • Tags Tags
    Accuracy Gravity
AI Thread Summary
The discussion revolves around determining the accuracy of the gravitational constant, -9.8 m/s², and how to incorporate it into calculations involving force (f=mg). It highlights that the accuracy of gravity can vary based on geological and altitude factors, but these are often not significant in controlled lab settings. Participants suggest using the accuracy of the measured mass when calculating total accuracy in the equation. The total accuracy is derived from the ratio of the mass's accuracy to its value, while the gravitational constant can be used as either -9.8 m/s² or 9.8 m/s², depending on the context. Understanding these nuances is crucial for accurate lab results.
StephenDoty
Messages
261
Reaction score
0
I am working on a lab and I need to know the accuacy of gravity=-9.8m/s/s? So I need the +- blank after the -9.8m/s/s. -9.8m/s/s +- _________m/s/s

Thank you

Stephen Doty
 
Physics news on Phys.org
Is there an accuracy error for gravity? Or do you just use the other number's accuracy error as the error of accuracy when doing f=mg. So the m has an accuracy is +-.02g so when doing f=mg what do I do for the accuracy error of gravity in the equation: total accuracy error= accuracy error of m/m + accuracy error of gravity/g.
 
Just use the accuracy level of things you actually measure. g is not constant, it depends on geology and altitude among other things. But those aren't likely to be variables in your lab experiment.
 
So for the accuracy equation for the f=mg : total accuracy= (.00002kg)/0.2704kg + 0/(9.8m/s/s)

From the equation: total accuracy = accuracy of m/m + accuracy of gravity/g

And for f=mg do I use (-9.8m/s/s) or (9.8m/s/s)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top