Homework Statement
Given the operator A = d/dx and B = x and the function f(x) = xe^(-ax)
evaluate : ABF(x) and BAF(x)
Do these operators commute (yes/No)
Homework Equations
[A,B]F(x) = ABF - BAF = 0 ; means they commuteThe Attempt at a Solution
[A,B]F(x) = ABF - BAF = 0
=d/dx(x^2e^-ax) -...
Hey Ray, please look at the updated attachment below. I have tried to edit the old one but don't see any button that would allow me do so. Thanks so much for your help in advance! :)
Thank you so much hallsoflvy for your reply. I am very grateful! I have attached a new picture to this email. I am looking for the value of A when the function is equal to zero. The final answer is sqrt (2/L). I want to understand the steps and how they got the final answer. Thanks so much in...
Homework Statement
Why must the PI be an average of two PKa's
Homework Equations
pI = (Pka1 + pKa2)/2
The Attempt at a Solution
the pI is the point when the net charge of a compound is zero.
I don't really know why it must be the average of two pKa's. Please I need help!
1) Please take a look at the attachment below in order to solve the problem. Using the figure at the right determine the values of i1, i2 and i3. Input your answers in the form "a.bc x 10^(y) A".
2) Using the correct results from #1, determine the following. For A and B you must also tell...
using substitution to solve i2 and i1, I get
i3 = -6/170 A ≈ -0.005381 A
i1 = ( -0.005381(15Ω+20Ω) + 3)/10 = 0.2811665
i2 = (4.5 - (-0.005381 (15Ω+20Ω))/(5Ω+12Ω)) = 0.2757844
using the information above, Please how do I calculate the current in the 10.0 resistor and How do i know the...
using substitution to solve i2 and i1, I get
i3 = -6/170 A ≈ -0.005381 A
i1 = ( -0.005381(15Ω+20Ω) + 3)/10 = 0.2811665
i2 = (4.5 - (-0.005381 (15Ω+20Ω))/(5Ω+12Ω)) = 0.2757844
using the information above, Please how do I calculate the current in the 10.0 resistor and How do i know the...
Thanks a lot for your reply.
I used the two loop equations to solve for i3, i2, and i1. Please Take a look at my work below...
i1(10Ω) - 9.0V + 6.0 V - i3(15Ω+20Ω) = 0
i1 = (i3(15Ω+20Ω) + 3)/10
i1 = 0.3 + i3(3.5) ......eqn 1
i3(15Ω+20Ω)+i2(5Ω+12Ω) - 4.5V = 0
i2 = (4.5 -...
1) Please take a look at the attachment below in order to solve the problem. Using the figure at the right determine the values of i1, i2 and i3. Input your answers in the form "a.bc x 10^(y) A".2) Using the correct results from #1, determine the following. For A and B you must also...
A potential difference of 50 mV is maintained between the ends of a 9.70 m length of wire whose cross section area is 25.0mm^2. The conductivity of the wire is 6.80 x 10^6 (ohm.m)^-1. Determine the rate at which the energy in the wire is transformed into kinetic to thermal energy.
2. Homework...