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Partial Pressure Help---Please Check The Partial Pressure of water vapor at T = 273 K is 6.11 mbar. Find the corresponding mass concentration of water vapor. Assume P = 1 atm. So.... ρH20 = (PH20*MH20/(R*T) ρH20= (6.11*10^2 Pa)*(18.01 g/mol)/(8.314*273 K) ρH20 = 11004.11/2269.722 =...

Thanks but otherwise the correct process right?

A typical global concentration of hydroxii radicals (OH) is about 10^6 moleculues/cm^3. What is the mixing ratio corresponding to this concentration at sea level P = 10^5 Pa and T=298 K? Using the previous equations derived earlier ndOH= (P*Av/R*T)*COH (ndOH*R*T/(P*Av) ) = COH (10^6...
4. Mass Concentration Problem

Perfect that matches with what I wrote it seems. Thanks. :)
5. Mass Concentration Problem

Much better now? Let's try this. Please check if right. Determine the mass concentration for N20 at a mixing ratio of 311 ppb at P = 1 atm, T= 298 K. My work nd N20= CN20 * ndair P*V = nmoles air* R * T (P*V)/(RT) = nmoles air (P*V)/(R*T) = N molecules air/Av = nmoles...
6. Mass Concentration Problem

Please check if right. Determine the mass concentration for N20 at a mixing ratio of 311 ppb at P = 1 atm, T= 298 K. My work #density of N20 = Concentration of N20 * #density air PV = #moles of air R T PV/RT = # moles of air PV/RT = Numberofmolecules/Avagardo # = #moles of air...
7. Physics Midterm Explanation Kirkoff Rule

Its the part c question....which states C) Suppose we made V = 0. Find the new current running through the 5.00 ohm resistor. So I just took the biggest loop and did it this way Sum of V = 0 So starting at the upper left hand corner of the circuit -3A(15 Ohms) + 50 V + I (5) = 0 -45 + 50...
8. Physics Midterm Explanation

Thank you! Sorry for being hard head! lol.
9. Physics Midterm Explanation

Can I use this instead.... q = Ceq Vo (1-e^(-t/RC) To find the charge Q?
10. Physics Midterm Explanation

Just making sure....the answer I posted was correct right?
11. Physics Midterm Explanation

Huh? I revised my post....
12. Physics Midterm Explanation

OHHHH...I think I understand now You use OHM's Law V = IR = IoR = Vo/R e (-t/taw) so that value x 5.00 E 3 ohms = ______ V V(t) = _____ V (from last line) (1-e(-t/RCeq) = ______ V Q = 3 E-6 F x ____ V (that I solved previously) = _________ C So I plug that into Q of the energy equation to...
13. Physics Midterm Explanation

Ok so you are saying this: Ceq = 1/C1 + 1/C2 = 1/12 E -6 + 1/4 E -6 but its the inverse because we are trying to find Ceq from 1/Ceq. So I got 3 E-6 F. I dont know what you mean to find the total charge?
14. Physics Midterm Explanation

Ok try to review for the finals but I need help in understanding this problem. So I can say that there is a resistance, 2 capacitors in series, and voltage source (Vo). Vo = 20.0 V R = 5.00 kilo ohm C1 = 12.0 micro F C2 = 4.0 micro F Switch is closed at t = 0. How much energy will...
15. Physics Midterm Question 3 Help

\frac{k}{\pi}\int \frac{-Q}{R^3}\mathrm{d}l = \frac{-k}{\pi R^2}\int {Q}\mathrm{d}l Where do I go from here? Because its supposed to be E = \frac{-2kQ}{\pi R^2}
16. Physics Midterm Question 3 Help

So lambda= Q/Pi R since the total Length is half the circumference of the circle is 1/2 (2pi R). dq = lambda dl dq = Q/Pi R dl E = integration of k dQ/R^2 E = k integration of Q/Pi R/ R^2 Is this right step so far?
17. Physics Midterm Question 3 Help

Integration of Kq/r^2?
18. Physics Midterm Question 3 Help

So Magnetic Flux = integral of E? Im confused about what formula to use on that question.
19. Physics Midterm Question 3 Help

It's half of a circular ring. Sorry I meant Electric Flux. Magnetic Flux = Integral of B dot Da = mew-naught x Ienclosed. So you are saying the E Field is not uniform? Is it because its not an enclosed circle?
20. Physics Midterm Question 3 Help

Ok, well I asked this question somewhere else and he said the same answer I wrote on my exam. But I got docked heavily for that. Ok well here's the problem: A negative charge - Q is distributed uniformly over a half circle with radius 'R'. P is at the center of the half circle. A) Show that...
21. Physics Midterm Help Part 2

Well, because I got 1.50 E -5 N for the Electrical Force. sigma = -3.0 E -8 C/m^2) Now if it was 2 εo q= F/E = (1.50 E -5 N)/(-3.0 E -8 C/m^2 /2(8.85 E -11 N/m) = -8.85 E -9 C Now if The E Field = σ / εo So ...q =F/E = (1.50 E -5 N)/ (-3.0 E -8 C/m^2 /(8.85 E -11 N/m) = -4.43 E -9...
22. Physics Midterm Help Part 2

so it just should be 1/2 of the value that I got before right? Oh is the E field zero inside the conductor because the E field go in and out of the conductor cancel out at the center?
23. Physics Midterm Help Part 2

Yes there is no conducting wire....it's a conducting plane of charge. Plus there was no radius given. But can you answer where the net electric field is zero in this problem? B/C I though it was at the center. But I got ding -10 points on the exam.
24. Physics Midterm Help Part 2

Yup a conducting plane of charge. I was trying to explain the diagram. Thanks though! :) I have one question is the net electric field zero at the center of the conducting plane?
25. Physics Midterm Help Part 2

So the E is E = sigma/2 Epslion knott?
26. Physics Midterm Help Part 2

Ok for part a....is the process right so since I already found what Electrical force is. I can just use F = qE ....which means q= F/E ...So I plug in F which I found already and E = Sigma/2 Eplison knott. Since we got the sigma value. But when I solved it I got q = -8.85 x 10^-9 C. Is that right?
27. Physics Midterm Help Part 2

Homework Statement A point charge 'q' hangs motionless from a mass less string at an angle, due to the forces of gravity and electrical attraction with a nearby infinite conducting plane of charge. Conducting wire sigma = -3.0 E -8 String is next to the conducting wire with a angle...
28. Physics Midterm Help

Homework Statement A two point charges (Qa and Qb) are placed along the y and x axis, as shown below: The Qa charge is placed 3 meters up from the y axis, which contains a -5.00 microcolumb charge. The Qb charge is placed 4 meters away from the x axis, which contains a 8.00...
29. Right hand Rule Question

Exactly but I still dont get it. Im doing this rule: "You can determine the direction of the magnetic force vector by using the right-hand rule as follows: point the fingers of your right hand in the direction of the velocity vector and then curl them around to point in the direction of the...
30. Right hand Rule Question

Well what I did was I just basically I curled it counterclockwise into my thumb. That's why I don see the force pointing to the east.