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 =...
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...
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...
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...
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...
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...
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?
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...
\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}
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?
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?
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...
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...
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?
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.
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?
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?
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...
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...
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...