FLu

Problem 10 1; -1 -1; 1 Problem 11 0000 0000 0010 0001 Anyone for Problem 4 and 5 please?

12 years ago

Anyone please?

12 years ago

No prob anynomous. Problem 13 a 1; 0 0; 5 Problem 12 a -1, 3 Would be nice if someone could figure out 4 and 5.

12 years ago

Thanks Anonymous!

12 years ago

RS (in Ohms): 100000 RL (in Ohms): 150000 ILmax (in Amps): 4.4e-5 ILmin (in Amps): 3.6e-5 Type all in at once, it does not accept one by one or errors!

12 years ago

Problem 1 a)No b)1 c)a; -b Problem 13 a) 1; 0 0; 5 b)Last tick for multiple question c)0 d)0 Anyone for Problem 5 though please?

12 years ago

Yes guys, Problem 5 please?

12 years ago

Click on FLu and you see the answers guys. Finally, anyone for Problem 5 please?

12 years ago

12)b fourth tick

12 years ago

Anyone for Problem 5?

12 years ago

12)b) Fourth Tick

12 years ago

Thanks, that's great!

12 years ago

Anonymous check circuit question, I have given the answer!

12 years ago

Hi Spanish friend, can you write in English? I had to translate it online and could not understand it. The answer is: 6.3 0.0137 6.3 Anyone for: Consider a deuteron in a cyclotron with field strength 0.5T. The deuteron is accelerated twice per rotation by...

12 years ago

Does anyone know this? Consider a deuteron in a cyclotron with field strength 0.5T. The deuteron is accelerated twice per rotation by a potential of V=25 kV. (a) If the radius of the cyclotron is 2 meter, what is the maximum energy of the deuteron? Expres...

12 years ago

Consider a deuteron in a cyclotron with field strength 0.5T. The deuteron is accelerated twice per rotation by a potential of V=25 kV. (a) If the radius of the cyclotron is 2 meter, what is the maximum energy of the deuteron? Express your answer in Joules...

12 years ago

Found c) Anyone for a) please?

12 years ago

Thanks abdrrahim! Anyone for the rest?

12 years ago

Thanks abdrrahim. However, mu_0 and I_0 is not allowed it says. can you help please?

12 years ago

Sorry abdrrahim did not realise it was solution for Problem 1. Do you have the others for Problem 2 please?

12 years ago

a)0.0000090020201102748 b)0.0025 c) First part -0.00358179 Anyone for second part of part c) Tesla per metre square?

12 years ago

Third and Last part of c) is 0 by the way. second part for c) Tesla per square meter please?

12 years ago

Anyone please?

12 years ago

Abdul Rahman do you have Problem 2 please?

12 years ago

Nobody for Problem 2?

12 years ago

Thanks, hmmmm. abdrahman mentioned that oen already above. Do you or anyone else have the other ones?

12 years ago

Thanks hmmmmm, what do you require? I need the second part of c) in problem 2 helmholtz hoils, do you have that please? The answers for the other, you can read in my profile, if you don't have it yet. Thanks.

12 years ago

Many thanks hmmmm! Have you by any chance found the second part (Helmholtz Coild)for Problem 3 c)? I have all others but that one with tesla in square meter is very hard.

12 years ago

Problem 2 is answered now Ruben, look at my profile. Do you have by any chance Problem 3 c) second part? tesla in square meter

12 years ago

That is easy Luis it is 18. Please let me know anyone who cracks second answer part c)? thanks

12 years ago

Thanks Luis, it is below! 4)b) (pi*B^2*L^4*omega)/R

12 years ago

All of the answers are wrong! Please help!

12 years ago

sorry answer a) seems right the rest appears wrong though. Anyone for b),c),d)?

12 years ago

Did you figure the above problem out Abdul? Thank

12 years ago

I got d) 8165 you multiply 12/0.09 and than apply what Henry provided for answer c) 12*126*5.4=8165 It seems the numbers are getting mixed up, can someone figure out b) and c) with it? thanks

12 years ago

I meant divide 12/0.09 and then multiply the result by 0.95 which is approx. 126 b) and c) anyone?

12 years ago

ANyone for Problem 2: Displacement Current a) please? b) 0 Open Switch on an RL Circuit b) and c) please?

12 years ago

Did somebody figure out b) and c) please?

12 years ago

Can somebody help with Problem: Displacement Current please? I thought the formula is: (1.25663706*10^-6*0.04*0.2*0.0705)/(2*pi*0.05^2) However, I am not getting the correctr answer.

12 years ago

Can somebody help with Problem: Displacement Current please? I thought the formula is: (1.25663706*10^-6*0.04*0.2*0.0705)/(2*pi*0.05^2) However, I am not getting the correctr answer.

12 years ago

Thanks Anonymous! Do you have by any chance the answer for Problem: Opening a Switch on an RL Circuit b) and c) I could figure out a)V_0/R_1 the other one seems more complicated. Can you help please?

12 years ago

Many thanks Anonymous! I got d) 100 Opening Switch on an RL Circuit: b) seems in a way easier as most of the letters are not accepted only R1, R2 and V0. Did somebody managed it yet?

12 years ago

dd, I have provided a and b already just have a look above. Anyone for c) please?

12 years ago

Sorry meant a and d. b) was provided by Anonymous. If anybody knows c), thanks in advance!

12 years ago

Anyone for c) guys please?

12 years ago

Thanks Supraconductor. I could not figure it out. Is it this: V_0*e^(R_1+R_2) Thanks in advance!

12 years ago

Anyone for Problem: Opening a Switch on an RL Circuit: c)? Thanks.

12 years ago

Anyone for Problem Opening a Switch on an RL Circuit c) please?

12 years ago

Great many thanks Anonymous!

12 years ago

Anonymous its the first. (1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))

12 years ago

It is this, thanks Andy! 5)a) ay = -1/sqrt(K) ax = -1/sqrt(K)

12 years ago

Thanks! Which of the following will increase the strength of a sample of 90% SiO2 - 10% MgO

12 years ago

Anonymous, can you provide formula because of different values please?

12 years ago

0.0287

12 years ago

a)0.2969 b)0.1347765

12 years ago

Have the same problem, how to calculate for A?

12 years ago

Thanks Phy: is this right? a) I = Sigma*A*omega 6*10^-4*4*pi*1*1*5 I don't seem to get it right. b) mu*I/(2*(z-R)) (6*10^-4)/(2*(1.92-1)) I am not sure, if I found I though. Please could someone help? thanks

12 years ago

Oh, I see. Anonymous or someone else, could you maybe provide an example with real numbers? That would also help others.

12 years ago

I could not get it right with that formula too. Anonymous, is there chance that you could give us a calculated example with the above numbers? We could trasfer it into ours, thanks.

12 years ago

Phy, did you by any chance manage Problems? Bainbridge mass spectrometer Magnetic Field of a current-carrying ribbon Second RL Circuit Magnetic Field of a loop: I got Bx and By but Bz could not figure out Bz. I have Problems 1 (RL Circuit) and 2.

12 years ago

Problem 1 as follows: a) I1:(R1+R2)/V I2:(R1+R2)/V I3:0 b) I1:V/R1 I2:0 I3:V/R1 Problem 2: I got right like this: if V right reads -0.5 then multiply by two and drop the minus, so a) 1 b) is the same value without minus, so b) 0.5 Anyone for the other pro...

12 years ago

Anyone?

12 years ago

Is value for mu not 1.25663706*10^-6? Is it different to mu_0 then? can you show us an example with the above numbers please?

12 years ago

Anonymous can you tell me what the value for A is then? Something does not add up. This calculation below looks odd to me: (6*10^-4*4*pi*1*1*5)*(omega/2*pi) Anyone did figure out yet?

12 years ago

Anonymous, the formula is wrong. Did you get the answer right by the way? If somebody got the answer could they provide it step by step with numbers though? thanks

12 years ago

Anonymous, the formula is wrong. Did you get the answer right by the way? If somebody got the answer could they provide it step by step with numbers though? thanks

12 years ago

oh I see, so there is a difference between mu and mu_0. I cannot find any values as well. could you just help me with calculation a) please? If my values are: sigma= 6x10−4C m−2, w= 5 rad⋅s−1 and R=1m Is A= 4*pi*r*r ? if not could you correct my values be...

12 years ago

you mean this is the formula for b) Anon?(mu/4*pi)/(2*(z-R)) Is mu same as mu_0? thanks could you also check if my procedure for a) is right Anon? thanks

12 years ago

Anon, is it this formula for b) then please? b)(mu*I/4*pi)/(2*(z-R)) and as Naseng said is the value for mu= 1.25663706*10^-6 if not what is it?

12 years ago

Anon, many thanks now a) I got but stuck with b) I would be forever greatful, if I can figure b) out with your help. a) 0.007 so that is my I b)(mu*I/4*pi)/(2*(z-R)) (1.25663706*10^-6*0.007/4*pi)/(2*(2.2-1) I followed your advice, but there seems to be an...

12 years ago

Thanks Anon but 1.25663706*10^-6 is 4*pi*10^-7, please check if you like. It still does not work, any other suggestion, am I missing a bracket somewhere?

12 years ago

hmmmm do not use for b) yet as there seems to be missing but I got a) right so can assure you the formula must be right.

12 years ago

No, but I think we are near there is something missing but cannot figure it out, maybe a bracket issue!? Anon any suggestion or even Anonymous? You are usually great help guys! thanks

12 years ago

Anon could you write the formula for b) please with the above numbers? would be very helpful.

12 years ago

Anon thanks, I thought it is (I*4*pi*10^(-7))/(2*(z-R)) did you not say mu is= 4pi * 10^(-7) ?

12 years ago

X10 there must be a problem with the grader as you suggested your friend got the right answer and a friend of mine got the right answer too with the provided formula by Anton.

12 years ago

In the circuit shown in the figure below, the switch closes at t=0, R=5.5 Ohm, ¦Å=9 V, L=0.08 H. (a) What are the currents (in A) through the two bottom branches at t=0+ (just after the switch is closed)? I1 and I2 ? (b) What are the currents (in A) throu...

12 years ago

X10 thanks for your help with Q6:((mu_0*I)/2*pi*w)*ln(1+(w/x)) (I = J*w*s) and convert cm to m. I am having issues with the numbers could you check if this is right please? If my values are: w=8cm; s=0.1cm; j=1A/m(squareroot) and x=22cm Is this formula co...

12 years ago

Thanks DOnny, did I convert it right? I want to be sure, as this is my last try. ((1.25663706*10^-6*(1*0.08*0.001))/2*pi*0.08)*ln(1+(0.08/0.22)) Thanks again.

12 years ago

DOnny my values are like this below: w=8cm; s=0.1cm; j=1A/m(squareroot) and x=22cm However, I am uncertain as Saga seems to got a wrong answer.

12 years ago

Sorry DOnny, have accidentally used your username. will let you know about Bz when I have figured it out. but have issues with the last problem still. Cannot figure out where the issue lies even though the numbers are converted to meters now.

12 years ago

Unfortunaley, can only use wolframalpha but that should work. Could you help me with figuring this problem out? my values are: w=8cm; s=0.1cm; j=1A/m(squareroot) and x=22cm Is this formula correct: ((1.25663706*10^-6*(1*0.08*0.001))/2*pi*0.08)*ln(1+(0.08/...

12 years ago

Thanks will try out!

12 years ago

DOnny you are a star! It worked after a huge struggle! I will try to figure Bz out and will let you know, if I get something. Thanks again!

12 years ago

Niznkl, thanks I got it now with DOnny's help. Which b part do you mean?

12 years ago

Niznkl, it did not work for me but a friend of mine solved it with this formula provided by Anton so it should work. X10 stated that the friend solved it too. However, there was an issue with ours. Use this, it should work and report back: remember I is t...

12 years ago

OH and before I forget the formula for Magnetic field of a loop last part Bz? Most of us had difficulties with that, so if you could provide us with the formula, I, DOnny, X10 and others would be greatful.

12 years ago

Niznkl, thanks I seem to have different values for I at least it is, I=0.75 and a=6 cm. COuld you give us the formula please so we can calculate it, or maybe a step by step guide? I could not figure it out. thanks! Anyone have formula for second RL circui...

12 years ago

Correction it is I=0.85 and a=6cm

12 years ago

Guys, I think the formula what X10 provided is for last problem on charged sphere not magnetic field of loop. The values can be found on it. X10 did you figure out Bz for magnetic field of a loop? and anyone for second RL circuit formula please?

12 years ago

Great thanks X10! For Problem 4, second RL Circuit question, could you help with formula? I know that first and last are 0 but the middle one I cannot get.

12 years ago

Anonymous, I think that is question 3 and was provided by X10. Just use the below formula, it works I tried. Q:3 for part a: B=E/v for part b: (L^2+h^2)/2L for part c: m=rqB/v

12 years ago

-0.71 mm Problem 3) 184 MPA Anyone for Problem 1 and 2 please?

12 years ago

Helpless it is, 184 MPA Do you have figured out Problem 1 and 2?

12 years ago

Yes, must have something to do with tolerance RORO, try -0.73, there was technical issue before. RORO did you get problem 1 and 2 please?

12 years ago

I have 1360 as answer. Do you have Problem 4: An LRC Circuit a) and b) please Anonymous?

12 years ago

Got it! Anyone for Lightly Damped Undriven Circuit (last problem) a) Express b in term of, if necessary, R, L, C. ? and b) ? I have problems with formula.

12 years ago

a) R/L second a) ? b) 0.000828 c) 0 0.00226 Did anybody get second question a)?Express b in term of, if necessary, R , L , C .

12 years ago

Anonymous, tried it out but it says rho_1 not allowed in answer. How did you manage?

12 years ago

Thanks Anonymous now it worked. Have you had luck with Problem 2?

12 years ago