What is it that you mean? The floating ground?
Yes I agree TINA has been giving me issues. Sometimes the outputs are different and all I do is delete a connection and redo the same connection and it changes everything. I found that out this morning.
Hello all! I hope I am posting this in the right place.
I am looking for some help on the following circuit. Basically I am doing a relay application using one of TI's chips. Below is the reference design from an app note along with my actual circuit modeled in TINA. The main function is that...
Homework Statement
Select all of the following which are possible combinations of Lz and θ for hydrogen atoms in a d state, where Lz is the z component of the angular momentum L, and θ is the angle between the +zaxis and the magnetic dipole moment µℓ due to the electron's orbital motion...
so this being said since the top connection is to the left the bottom is to the right?
& I guess what I am Really confused at is which one is positive. The tutor explained to me both the connected terminals where positive on top and negative on the bottom. I am more lost on determining which...
Chegg has this. This because the power A > 0 so it is being delivered power to A and it flows from negative to positive?
Is my assumption correct?
thank you
Homework Statement
I am having trouble understanding PSC. Please help on this question. Teach me like I am 5. My tutor couldn't break it to me.
(SEE ATTACHED) and find whether the power is being given or taken
Homework Equations
p=vi
p=-vi
The Attempt at a Solution
I can't...
I had done so much work I didnt want to type it and was going to upload a picture, But indeed I figured it out.. Using U sub with U = xy^2 du = 2xy and x was a constant so it was factored out leaving everything peachy!
Homework Statement
Attached below
Homework Equations
The Attempt at a Solution
So I cannot figure this out. Would this be integration by parts? or by substitution.. It provides me with an answer but no reasoning behind it and I cannot figure it out =/
Well I'm pretty retarded lol
But what about the greater than or equal to signs? And why weren't they reversed? Like from elementary you learn to flip them if it's anything with a negative number.
Homework Statement
See attached image
Homework Equations
they are provided in the image. I see the domain is used
The Attempt at a Solution
I can't figure out how they went from step one to step 2. where does the -81 come from and the 9 multiplied to the function in the middle. I...
On the picture it depicts it. An in the paragraph starting with "some of the electrons" sentence says they are unbalanced. I'm just a little confused on it
Can anyone explain this excerpt to me? I'm lost on how the positive and negative charges push each other away instead of attract.
Thanks!
P.s
I think this goes in this forum.
Homework Statement
Please see the picture.
Homework Equations
How do they get this derivation?
The Attempt at a Solution
I know that lambda = 2L/n and f = nV/2L so I can somewhat rearrange it myself, but I do not know where the square root is coming from.
ahhh i see so the first value i get for x will either be added or subtracted from the half period or however you want to phrase it. so if its 39.3° then 90°-39.3° = 50.7°
right it crosses the graph twice. just before the peak like he drew it, but how is that figured out? i know it is between half a period, or (pi) because it repeats itself after that. and in radians its .68.. help me figure this out without telling me! lol i really want to deduce it on my own
Homework Statement
Sin(2x)=.98
Homework Equations
I know you take arcsin of both sides and divide by 2 to get the answer which is 39, but where does this guy get 50 from? I would have assumed it occurred again at 2*39.. not 50. it is an online lecture.
The Attempt at a Solution...
I see where you are coming from! just makes no sense to me to square it given I had all the variables in the experiment that we did in class. Maybe for future reference know this is something great to remember. Thanks Jack!
Homework Statement
I am having an issue with answering number 4 in the attached image.
Homework Equations
Relevant equations are given in question 3.
The Attempt at a Solution
Squaring the equation would only make it easier to solve for any of the other variables, or the period...
Homework Statement
Find Beta of <60,-50,40>
Homework Equations
Cosine β = Ay/lAl
The Attempt at a Solution
lAl = 87.74964387
So its arccos(-50/87.75) = 124.62
express to 2 significant figures and it keeps telling me I am wrong. I rounded to 125. kept at 124.62.. etc.
well 180 in a line.. so 180 - 105 = 75.. for the "quadrant III" let's say.. opposite is 75 total.. but you have 30 given. so that makes the extra angle left 45. so the alt interior angle tells me that there will be a 45, 30, 105 triangle
Homework Statement
Attached below
Homework Equations
I am having a problem because the solutions manual does the parallelogram method and stupid me I cannot draw the parallelogram correctly. Specifically my drawings are off a bit. and it throws me off every time. the last one I had...
ω=2 because 2 squared = 4. so just to make sure I am doing -(w*w)(x)
I picked positive 5x before because I was an idiot and thought the negative was attached to the ω value.
the position function for a particle in SHM is x(t) = Xmcos(ωt+ø)
differentiating it you get velocity. and again you get a(t) = -ω^{2}XmCos(ωt+ø) ... = a(t) = -ω^{2}x(t)
being that it needs to have the equation form to fit, yes I say it is reasonable. a(t) = -ω^{2}x(t).. that being said -4x is the only option that seems to fit.
based on what you're saying now I think it is either 5x or -4x. the force is directly proportional to the acceleration. I think I am missing the final key point
I have the text and I now remember the (-) sign in the equation. Completely my mistake. I have 2 books I am trying to learn from so forgive me if I forget something.
So, if I understand this completely, I was wrong because there is no x^2 in regards to the acceleration. it would be the -4x option because it seems to be closely related to what we just proved. I think I was thinking of ω^{2}
The gravity would be pulling it down and the force would be opposite.
Or with a block the spring is applying the force if it's on a table both ways. Acceleration will be applied opposite of the motion. So the force is given by the spring