Is it possible to measure the PD of photocells using just a voltmeter?

  • Thread starter Thread starter Iamphoebe
  • Start date Start date
Iamphoebe
Messages
3
Reaction score
0
I've looked everywhere for information on photocells and a distance-output equation, but have found nothing.. Anyone got any ideas on this?

edit: forget it, sorry, i have found another thread on this forum . thanks guys. :D
 
Last edited:
Physics news on Phys.org
ok just one thing, the use of a voltmeter alone to measure the pd, i gather from your responses, is not possible. why?
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top