If you have 2 building with a rope connected to each end of the buildings that is 10 feet long and has a 5 foot dip in the middle of the building how far apart are the two buldings?
133553
the buildings are touching! If there is a 5 foot dip in the middle, then when it is attached to both ends of the building, then there is no room for any space between the buildings, because the rope folds over.
55639
I think the answer should be\n\n2 * 10/(1-4/5) - 10  = 90 ft\n\nas the new height would be 4/5 of the immediate previous height, so R = 4/5.\n\nand the sequence would be 10, 8, 32/5 ....
401135
The secret to this question is to remember that the force exerted on the top rope (tension) is based on the mass of both boxes (20 kg) since it is actually holding both to the ceiling. The tension of the bottom rope will only be based on the mass of the bottom box.\n\nHope this helps. Good luck!
410203
And find the slope of the line, and extend a line of the same slope and length from R to find the other endpoint.\n\na) slope = (-12 - 22)/(52 -12) = -30/40 = -3/4. Extending a line with the same slope and legth in the opposite direction can be done by going -40 units along x and 30 units along y. The other endpoint will then be (12 - 40, 22 -(-30)) or (-28, 52).\nb) For this we find the distance between the two endpoints. D = sqrt((52-(-28))^2 + (-12-52^2)) = sqrt(80^2 + 60^2) = sqrt(6400 + 3600)  = sqrt(100^2) = 10. The circle has a diameter of 10.
440292
v = rω\n=10 * 7π\n≈ 219.9 ft/s
640913
This is not the correct answer.  While the shortest between two points is a straight line, the distance between 4 points is not.  Actually, there is a principal which is referred to as Steiner Points.  This is applicable in this situation.  Two Steiner points (P) are placed inbtween points ABC and D.\n\nImaging a Y, except with a triangle, similar to the upper portion of the Y attached to the bottom of the Y.  Essentially, you would have a triangle connecting points A and B and one connecting C and D.  However, their height is only 2.9 Miles as opposed to 5 miles.  The remaining 4.2 mile gap would be connected with a straight line between each triangles point.\n\nThe two diagonal lines of the triangles would be 5.78 miles, or 23.12 miles for all four.  Now you add in the 4.2 mile straight line, for a total ROAD distance of 27.32 Miles.
384085
w1=width of 1st rect.\nL1=length of 1st rect\nA1=area of 1st rect\nw2=width of 2nd rect\nL2=length of 2nd rect\nA2=area of 2nd rect\nw1=x\nL1=4x\nA1=length*width\n=4*(square of x)\nw2=x+2\nL2=4x+5\nA2=(x+2)(4x+5)\n=x(4x+5)+2(4x+5)\n=4*(square of x)+5x+8x+10\n=4*(square of x)+13x+10\n\nAccording to the given condition\nA1+270=A2\n4*(square of x)+270=4*(square of x)+13x+10\n270=13x+10                //4*(square of x) gets cancelled\n13x=260\nx=20\ntherefore width of 1st rect=20cm\nlenght of 1st rect=20*4=80cm
100505
Maybe I'm not understanding the problem correctly, but if "There is only space for the pature 2 be @ most 100ft" means that no side can be longer than 100 feet then I think you may have written the question down incorrectly.\n\nIf each side has to be less than or equal to 100 feet, then let's say that each side *is* 100 feet in length. There are 4 sides, so the perimeter is 4 times 100 feet = 400 feet. You say you have 500 feet of fencing. This is more than enough to surround the maximum-sized pasture allowed. So what's the problem?\n\nThe only two linear equations you need to worry about in this case are:\nwidth <= 100\nlength <= 100\n\nChoose any values for the width and length that satisfy these constraints,\nsay the width = 75 and the length = 50 or the width = 2 and the length = 89\nand you have solutions to your question.\n\nOften these questions ask for the maximum area that it is possible to enclose. In this case there is only one solution: width = 100, length = 100 therefore the area enclosed = 10,000 square feet and the amount of fencing used is 400 feet.
410295
A is moving from Q at 2 ft/s (= Va)\n(Velocity of B towards Q = Vb)\n\nSo Vrope = Va /cosα (α = angle rope makes with horizontal at A)\nNow if AQ = 5 then hypotenuse (= length A to pulley) = 13 (as 5² + 12² = 13²)\n\nSo Vrope = 2/(5/13) = 26/5 ft/s\n\nNow length of rope from pulley to B = 39 - 13 = 26 ft\nBy Pythagoras' Rule length QB = √(26² - 12²) = 2√(133)\n\nVb = Vrope * cosβ (β = angle rope makes with horizontal at B)\n= 26/5*2√(133)/26\n= 2√(133)/5\n≈ 4.61 ft/s
14632
here you assume there is almost no point of intersection...or at least that it is exactly a point therefor no measureable space so you must determine the two circumfrences...use the equation\nc=pi * diameter\nc1 = pi *6\nc2 = pi * 18\n\ntotal distance is 24pi\n\nhope that helps!\n\nactually, that is a good point, when i first read it i assumed that we would want to put the rope around the circle...meaning that we would have the rope touch the circle at all times...the problem with stating that there is a line from one circle to the other will be slightly inaccurate, the line from the two circles will be different from the horizontal by several degrees, that would be the need for teh trig function.  I would say that for this point in the school year for a high school geometry class they probably havnt gotten to trig yet...\n\nhope this helps!!
527006
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
468074
let the height of cliff be h\nthen \ndistance of first ship =x\ndistance of first ship =x+500\nnow \n1.h/x=tan 24\n2.h/(x+500)=tan 15\nfrom 1\nx=hcot24\nfrom 2\nx=hcot15-500\ntherefore\nh(cot15-cot24)=500\nh=336.47
211121
Trigonometry is hard to do in you head, so start out by making a drawing. You have two buildings A (tall) and B (short) - Draw them far apart, not like this:\n\nA         B\n|         \n|         \n|    \n|         |\n|         |\n|         |\n|         |\n\nNow draw one line from the top of A to the top of B - this is the angle of depression of the top and it is 46 degrees. That is, if you drew a straight, horizontal line across the top of building A, the angle between that line and the line joining the top of A to the top of B is 46 degress.\n\nSimilarly, draw a line from the top of A to the *bottom* of B - this is the angle of depression of the bottom and it is 77 degrees. That is, the angle between the horizontal line across the top of A, and the line between top of A and bottom of B is 77 degrees.\n\nNow can you find out the angle between the line from A-top to B-top, and the line from A-top to B-bottom? (It's very important for you to draw this!) Let's put this value aside for a minute. \n\nLooking at your drawing, can you see that you know the value for *part* of the angle between the horizontal line across A's top and length of building A? What is that part you know? Now, see how the line across A's top and length of A make a right angle? So how many degrees in a right angle? So if you know the total of that angle, and part of that angle, how do you get the other part? **Notice that this part we're trying to get is actually the angle between the line A-top to *B*-bottom, and the line A-top to *A*-bottom. \n\nYou should also notice that now we have two triangles:\nTriangle 1: A-top to B-bottom to A-bottom. Solve this triangle for line A-bottom to B-bottom for distance between the two buildings.\nTriangle 2: A-top to B-top to B-bottom. Solve this triangle for line B-top to B-bottom for height of building. \n\nDon't forget, we have two important angles that will help solve these two triangles!
432460
a) Assume the following:\n\nAngles a= 10 deg, b= 80 Deg, c=90 deg \nHypotenuse length = 2.5 m\nheight = 2.5 * cos 80 Deg= .434 m\nlength = 2.5 * sin 80 Deg= 2.462 m\nacceleration = "g" * Sin 10 Degree
356498
Everyone here, including the asker, is wrong.\n\nThe long rope will carry a greater load than the short one if the initial tension in each is the same.\n\nReason:  The long rope will DEFLECT more than the short, and the angle it makes with the horizontal will therefore be greater.  The greater the angle, the more upward load the rope will carry.\n\nTorque has NOTHING to do with this problem.  The nature of a rope is that is can ONLY support TENSION.  No bending (torque) allowed.\n\nI'll supply the equations in a while; I started scribbling them down, but they got a little hairy, and I wanted more time to sort them out so they're readable.  I like this problem; it's just that you assumed a particular answer was right a priori..........
340256
find the distance of the b1 and b2 then plug them in
161513
You get K by knowing that the parabola is centered in the middle of the span (where the sag is lowest). So\n\nk= y/x^2  = 143/640^2  = 3.5x10^-4\n\nFrom there use the path length equation to get the length of the cable. When you have that, multiply by 1.05 and work backwards for (b)