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Ling
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Ling
Questions (8)
Cn u plz gv explaination 2 d solutions bit by bit in brief? Cz i really dnt understand hw 2 apply d formula 2 solve it when i
1 answer
433 views
An analysis on East London By Matthew Arnold please??
'Twas August, and the fierce sun overhead Smote on the squalid streets of
9 answers
1,153 views
Can you create a java program that will make a CPU SCHEDULER w/ First Come First Serve,Round Robin,Priority
2 answers
1,007 views
Consider the gold-platinum system which has a value of delta w of -4250 J/mole;
(a) Calculate the enthalpy of mixing for a
7 answers
602 views
Answer the following questions by referring to the Cu-Al phase diagram shown below.
(a) What is the lowest temperature at which a
10 answers
535 views
(a) Silicon (50 g) has been doped with 27 mg of antimony (Sb). Determine the concentration of free charge carriers
10 answers
456 views
A photon interacts with a ground state electron in a hydrogen atom and is absorbed. The electron is ejected from the atom and
6 answers
480 views
(a) A pressure vessel contains a large volume of CO2 gas at 10 atm pressure. A membrane composed of a poly(ether ketone) polymer
28 answers
585 views
Answers (41)
YOU WANT COOKIE??!!
He meant precursor Ms.Sue
All winters answers are correct
All Stacie answers are correct
Ayo 😂😂 tf y'all dumb funny ðŸ˜ðŸ˜
Thanks got 100% B A B A
Y'all really weird for that like ard y'all act like Ben the only one who do it and stop ick eating off of connexus
Thanks 100%
Yes winter was all correct 😂
@gwen is correct tf this is basic stuff
Number 13 is wrong it's D because it decreased
Thank you
Is 39.2 in eV?
Thanks
a) 6.22*10^18 b)wavenumber= Eg / hc Eg(Silicon band energy)= 1.11 eV (to joules) h=6.6*10^-34 c=3*10^8 wavenumber= Eg*1.6*10^-19 / hc = 896969.69
Thank you for the explanation.
a) 6.22*10^18 b)wavenumber= Eg / hc Eg(Silicon band energy)= 1.11 eV (to joules) h=6.6*10^-34 c=3*10^8 wavenumber= Eg*1.6*10^-19 / hc = 896969.69
a) 6.22*10^18 b)wavenumber= Eg / hc Eg(Silicon band energy)= 1.11 eV (to joules) h=6.6*10^-34 c=3*10^8 wavenumber= Eg*1.6*10^-19 / hc = 896969.69
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
First find: E = P^2/2Me + E(first ionization) P= h / BroglieWavelength E(first ionization)=21.7*10^-19 Me= 9.1*10^-31 h= 6.626*10^-34 Once E is found the find frequency: frequency(hz)= E / h
Thank you
B) is not 45 and A) is not 18
How to input the data into the formula?
same here, how to insert the data into the formula?
Could someone insert the data into the formula?
Step 1: (0.5*7715.0 - 7715.0)/ (165.0 - 7715.0)= erf(ξ) Step 2: erf(ξ < 0.6), use the approximation erf ξ = ξ Step 3: t= d^2/4*D*ξ^2 d=61.5 x 10-2 cm D=3.091 x 10-7 cm2/s Step 4: t in sec -> hours Conversion seconds to hours is total/(60x60)
Step 1: (0.5*6910.0 - 6910.0)/ (379.0 - 6910.0)= erf(ξ) Step 2: erf(ξ < 0.6), use the approximation erf ξ = ξ Step 3: t= d^2/4*D*ξ^2 d=13.0 x 10-2 cm D=3.091 x 10-7 cm2/s Step 4: t in sec -> hours Conversion seconds to hours is total/(60x60)
Had wrong for question two with I
What is b and c?
What is b) ?
Had 3.24888828e-07 but place 777.78 and had wrong answer. What when wrong?
Thank you
I meant Arnold's advice to the reader/audience. But to the above question i think he does agree from the last three lines. "not with lost toil" says a lot about how Arnold think will happen to people if they have faith.
Thank you so much! One last thing. Is he confining the hope to just religion and God? or can hope/faith be found in other things too?
Well the old english is killing me. But so far i think that Arnold using the conditions of london during the 1800s to show how important faith and hope is. The preacher in the poem found strength in religion and God, and Arnold is saying that if one can
