At the Introductory Physics page where you posted this, there is a sticky at the top with different formulas in there. The link is <https://www.physicsforums.com/showthread.php?t=110015> and the formula you need is in there. It's in the second post that Doc Al posted under "General".
You have...
It's been awhile since I've done any of this so somebody correct me if otherwise, but for west of north and for above the horizontal, you should not be using any distance as the hypotenuse.
You will need two different triangles for each situation which you did, but the hypotenuse is not...
I had this problem or almost exact one back in high school. I found it easy to draw a picture, convert each vector into vertical and horizontal components...
I think I get it, so you just calculated the hypotenuse of the triangle to be 60 as well? Which you didn't really need to do, but it's good to double check too. But yes, the tree I got was 30.02m >> 30m.
You should get an answer of 30.02m which makes sense. If the shadow was the same as the height of the tree, that would suggest the angle is 45 (middle of 0-90). If the shadow was shorter than the tree, it would suggest the sun be more overtop of the tree, giving an angle closer to 90...
You were right with the 52, I managed to put a crappy diagram there, but O is the sun, and that line ' - ' is the line that makes an angle of 30 degrees with the sun and the horizontal. So you say the angle should be in the corner A, but does the diagram help you to determine if you are right...
Best way I find to do most questions like this is to draw pictures. So, this will end up being another triangle.
So start with a tree, so a line going up, now then I drew the "shadow", so another line at the base of the tree going to the right.
Now draw a line connecting the lines to...
What did you get for the angle, I just want to see if you managed to get it. The angle where line D meets the horizontal? I'm hoping I did it right but you should get... 60.9 degrees... I think haha.
You have to specify sometimes too on your diagram, or if you don't draw one with words. Now...
Ah k, so you have 2 angles, one up where line D meets line A (vertical line) and we'll call this angle phi. You also have an angle where line D meets line B (horizontal line and we'll call this angle theta. Then a 90 degree angle where A meets B. Now you can use either angle phi or theta...
Should be actually inverse tan I believe, along with the options of inverse cos, and inverse sin
tan x = opposite/adjacent
x = inv.tan (opp/adj)
or...
x = inv.cos (adj/hyp)
or...
x = inv.sin (opp/hyp)
K, well i'll attempt this, draw the lines out on a paper. So draw one line from the start point (call this anything, I'll say C for explaining purposes) going to the right (east) 50 so let's say 5cm (even though in this case the drawing doesn't need to be exact). Now from there, draw a line...
Basically ya, this is what I understood from the question. Hopefully that's what you wanted to do. Then you may* need to find the angle of displacement.
Alright, well the vertical (positive or negative) would be the north and south ones. So let's say up (north) is positive. How far from the start position are you if you go 40 units south, then go 130 units north? (Ignore the east direction for now)
I'm pretty out of it (tired), but I'll see what I can explain. From my understanding, this is dealing with vectors. Now it's hard to tell without a diagram or more knowledge of the question but for each direction, break it down into vertical and horizontal directions, then add up all the...
I was wondering if anyone knew of links to different websites that stated the coefficients of resistance for different metals.
Mainly copper, and steel equipment. I am working in a mill and my supervisor asked me to find out these values. There is copper cable running as ground and...
I recognized there was a pattern across, but at first didn't see there was a pattern going down. Number 13 is still bugging me a bit because it's something easy that is being overlooked I'm sure lol. If you get a response to that question you should let me know.
This relates to this problem at the mill. There were readings taken from around the mill, on the PDCs and the breakers/equipment relating to each PDC. Now what I was looking for by posting this PDF attachment was anything odd from the readings.
The only thing I really noticed was that in...
It's been a few years, so I am not really sure on this. Since it's oriented outwards, and there's 4 sides, the sides opposite each other would cancel each other out... (I think, but hopefully I get corrected if not) so the only surface that contributes is the top.
Again, I would wait for...
Alright thanks.
If I took a different approach and took the Ohms/(1000feet) value of the cable, and multiplied that by the length (600 ft) then multiply that by the current... would that be another method?
= .6282 ohms/Kft * 600ft
= .37692 ohms
V = IR
= 12A * .37692 ohms
=...
I read briefly over most of the posts, but from what I collected, you have to walk into and exit the house but can do this several times. You cannot use a door twice.
Now if you have to enter, you must exit the house. Since there is 5 doors on the exterior of the house, you enter... exit...
Ya, I ignored that for a second when doing the calculation and got the same 0.005...V that you did, units cancelled. I'll wait for somebody else's reply because I am a bit confused now too haha. Do you have a correct answer for that?
It's been a long time since I've done any questions like this, but the equation "V=B/T * A * N" you use but you do not use the 'T' anywhere. You have "V = 8.15e-3 * (.218/2)^2 * pi * 18 turns" but unless I'm missing something, I don't see 'T' in there or a value of 'T' given in the question.
I was just looking for some insight on this problem. It is not a homework problem, my boss just asked me to figure it out.
Now, there is 120V supply, with 600 feet of cable (copper) running to a load, and 600 ft neutral coming back to the supply. Ignoring the resistance of the load, he asked...
50/50 chance there, Try again :P. If you have a smaller current flowing, less electrons (therefore: charge) would be flowing. Making sense... somewhat?
Should be...
x = \frac{u^{2} + 1}{2}
Shouldn't it?
I managed to get the integral with 2u on the bottom and looking the same, but let me know how you're making out.