The reading on the voltmeter is 0.
I figure the potential drop through line AB must be the same as the potential drop through AC (since voltmeter is zero). As such, the 10 Ohm resistor has the same as the effective resistance of the parallel branch.
1/10 = 1/8 + 1/R
1/R = -0.025
R = -40 Ohms...
Indeed, but if I directly convert 1672 MeV/c^2 using the fundamental charge:
(1672 * 10^6 eV/c^2) * 1.609 J/eV, I get 2.68 * 10^-10 J/c^2. This is kilograms.
Same wrong answer. I'm missing something here. I have to divide again by c^2 and I have no idea why.
(1672 MeV/c^2) * c^2 = 1.505 * 10^20 MeV = 1.505 * 10^26 eV = 2.41 * 10^7 J
Since E = mc^2, m = E/c^2
Therefore, m = 2.41 * 10^7 / (3 * 10^8)^2 = 2.68 * 10^-10 kg
But the answer is 2.97 * 10^-27 kg
Help! What is wrong with my logic?
Indeed, the message is sent via another rocket of some sort (I made the question up). It's still nice to see your thinking behind a carrier wave though!
Thanks to both of you for taking the time to explain.
u is the speed of the message relative to the station
v is the speed of the spaceship relative to the station
u’ is the speed of the message relative to the spaceship
u=(-0.5c+0.7c)/(1+((-0.5c)(0.7c))/c^2 )
=0.2c/0.65=0.308c
This just seems way too high, and I'm not sure if I'm doing it...
I removed the coefficient of 1/2 before integrating. So I had:
1/2 integral[(1/x-1)]
= 1/2 ln(x-1) + C
Using u substitution without removing the coefficient yields
1/2 ln(2x-2) + C
I get how it works both ways, and my answer must be wrong, but what exactly is causing this conflict? Am I not...