actually that is exactly what your talking about... i just drew it out with the source and a resistor in series and did KVL and yes i totally understand now... but basically this is not giving true power dissipated by the circuit is it not... its now including power dissipated by the power...
im not 100% sure what you mean by that but im assuming you mean... an ideal DC power source has 0 losses and a non-ideal power source does have losses?
During a Lab we had a simple purely resistive circuit hooked up to a 10 VDC power supply with an adjustable voltage output. In order to determine the power dissipated by the circuit we were asked to determine the source voltage while the circuit was DISCONNECTED... we were not told why... In the...
Oh excellent, so judging by what you wrote to continue past where I left off it involves a double integral? if this is the case I have not learned the double integral yet this semester which makes more sense why I as so stuck ..
Can someone please help me work through this problem I've spent over an hour on this trying to figure out what to do.. heres the question
A nonuniform electric field is given by the expression E = ay^i + bz^j + cx^k,
where a, b, and c are constants. Determine the electric flux through a...
I understand that part, its the actual meaning of the symbol dQ... or dx for that matter that im looking for
what im finding so far is that dQ could mean an infinitly small portion of a larger charge Q
I'm taking a calculus based physics class this year and I've had a few issues getting reliable information on the following notation
the following equation I = \frac{dQ}{dt}
the previous equatin represents the instantaneous current in a conductor.
what exactly is the term dQ or dt...
excellent thank you,
I thought that was the method but i'm still running into trouble expanding cos(x + 25) and dividing to clear... when i expand i get
cosx / (cos(x))(cos(25)) - (sin(x))(sin(25)) = 6/7
i dont see how dividing that by cos(x) now accomplishes anything?
where am...
Homework Statement
i'm having troubles solving this problem in physics involving dot product rule
Vector B is 5m and 60 degress above the x axis
A has the same magnitude as C and C is has 25 more degrees than A ..
find magnitude of A..
B dot C = 35
B dot A = 30
this all we're...
Homework Statement
Quick question I can visualise but not determine mathematically..
The question in my text says:
Determine the MINIMUM size of a flat vertical mirror that a person 178 cm can see his whole image. It suggest that a ray diagram will help.
The Attempt at a Solution...
okay nice! thats logic I've been messing around with as well and it seemed everything was going good and then I thought...well if T1 and T2 are 90 degrees and have 0 horizontal component then we assume 89.999999 degrees etc... can the same not be said for the load pulling downward giving it a...
is this not a joint sum of forces (note T1 and T2 are seperated for clarity, you can think of them as attaching to the same point)? sorry to keep bugging you on this one. I really want to make sense of it and have not heard the term joint equilibrium before but from what I gathered on what I...
very good answer thanks so much for looking at my problem... There was some clarification issues on my behalf but you covered each situation perfectly. The initial question I posted "attached to seperate anchors" was misleading because i meant seperate anchors for connection to a wall or...
thats also a good way of looking at it. what makes T1 and T2 unequal... well im not sure in this case.
the two ropes are parallel with each other... and attached to the load at the same spot. one rope is longer than the other. the forces are only in the Y axis.
I cannot think of any other...
There has to be a way that physics can solve this sort of a question short of sticking a load cell on the ropes to determine that the load would in fact be split equally.... I'm definitely missing something here. maybe its calculus I have no idea. but if i take those two ropes that form an angle...
yes third post was a bad example i was looking at a symmetrical system at that point..
so what dictates a perfectly symmetrical system in a situation where the two ropes a perfectly parallel with each other... if one rope is longer than the other.. yet they are parallel the force should still...
if you focus in right at the box no you cannot both ropes would leave the connection at the box from roughly the same point perfectly parallel..
so i can see the force would have the exact same direction. but how do i know the magnitudes would be equal
well say for example rope1 was is 10ft and rope 2 is 4 feet.. so they are hung from diffrent heights but in the end they attach to the load at exactly the same point parrallel to each other.. the lengths are different so you can definitely see a difference between rope 1 and rope 2 but .. the...
As per the diagrams attached on this reply.
Situation a.) shows what i am trying to figure out.. and situation b.) shows a more complex situation that its clear to see that the two horizontal components must equal each other since the object is not accelerating the sum of forces on the x access...
it just seems odd to me that if i were to take that exact same question and instead of having each rope parallel to each other i make an angle of 140 degrees between the two.. i can tell everything. Since it adds a horizontal component and the system is at equilibrium TIx must equal T2x and...
Homework Statement
This will probabaly seem like a very simple question and maybe I'm wayy over thinking it.
This goes back to forces on an anchor system.. lets say I have two pieces of rope holding up a 70 kg load.. each rope is attached to a seperate anchor.. the two ropes are perfectly...
So I found a report that pretty much sums up everything I've been trying to create only they do a wayy better job and it's part of an actual course with the title (go figure) "the physics of rock climbing"
definitely check this out...
Hey, back again ended up working all the way through the long weekend, but thats great to hear you get out into the mountains. keep it up... it really is a great stress reliever. to me it's what life is all about. Anyways I revisted this question with some new knowledge (my spring skills...
hey! thanks! thought you might find that interesting... yes I've heard of the american triangle, never used it. not a good situation at all.. gotta love the physics involed with the whole sport of climbing!
Thanks again for your well put answers, it's begining to clear things up a lot more. I will mess around with some questions involving conservation of energy and spring potential energy which will hopefully clear that aspect up for me.. the rope stretch makes this quite interesting.
Anyways in...
awesome I love your replies, So im guessing this is the same principle used in a haul system (pulley system). Hypothetically speaking if the medium the anchor was attached to was able to move, and we had the rope end where the belayer is attached to completely fixed then a person could move a...
Thanks very much for your help, I should tell you im not useing what is said on this forum to determine how I will build my anchor, just very curious. I've been climbing for many years now and feel very solid, however I just read that the way I've been belaying a second climber on larger...
Homework Statement
So after determining the way an anchor setup will handle the high tension forces created when at different angles, here is the next step in the question.
attached is a rough diagram that can be refered to for some clarity (hopefully).
So assuming we have a 68kg climber...