i have a math problem that involves finding the distance around 3 pulleys where do i begin?
146387
well question is find the dis around pulleys..take the 1st pulley find its half circumfernce. now go to 3rd. find its half circumfernce.now add both with 2*central distance of 1st and 3rd....\nu havent given the complete ques but can use simillar method to solve...
833623
perpendicular distance s the shortest\nlet the pont be (x,y)\ncentre of the circle=(1,2)\nequation of the line joining (1,2)and (-2,-2)\n(y-2)/(-2-2)=(x-1)/(-2-1)\n-3(y-2)=-4(x-1)\n4x-3y+2=0\nsolve this with the equation of the circle\nto get the coordinates 'x' and'y'
948476
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
1129609
A pulley is a chain, belt or rope wrapped around a wheel. The mechanical advantage of a pulley system is approximately equal to the amount of supporting ropes or strands. \n\nTherefor, if you had a mass of 60kg and wanted to lift it using two supporting ropes, you would have mechanical advantage (MA) of 2. The mass will feel like one half of what it really is. When lifted with the help of the pulley system your 60kg would only feel like 30kg. Thus the effort force equals 30kg. \n\nHere are a couple Of websites that explain this more.\n\nhttp://www.technologystudent.com/gears1/pulley7.htm\n\nhttp://www.swe.org/iac/lp/pulley_03.html
149565
cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
648691
Divide all 9 balls into 3 groups of 3 balls.\n\n1) weigh any 2 groups:\n         - if they are equal, the crooked ball is in the 3rd group\n         - if not, you have found the group where the ball is\n\n2) repeat the same experiment with that group until you find the ball
951341
The linear speed of the cable is\n181/17.2 ft/sec\n\none revolution is equal to the circumference of the pulley or 7.7 *pi ft/revolution\n\nso the angular speed is \n(181/17.2)/(7.7*pi) rev/sec\n=.435 rev/sec\n\nj
252925
The two simplest pulley systems demonstrate the principles of their operation.  First if there is a single pulley overhead and you pull down to lift a load, you must pull with the same force as the weight of the load because oonly one rope pulls up on the load.  100 pound load = 100 pound pull.\n\nSecond, if the load is below you and you pull up on a rope connected to your own level through a pulley, you pull with a force one half the weight of the load because two ropes are pulling now up on the load.  100 pound load = 50 pounds pull + 50 pounds on fixed rope.\n\nThe overhead pulley required you to pull the rope say 10 feet to lift the load 10 feet; but the pulley below you required you to pull the rope 20 feet to lift the load 10 feet.\n\nYou can calculate more complicated pulley systems either by counting the number of ropes that are under tension that actually lift the load and dividing the load by that number or you can determine the length of rope you must pull to raise the load a given distance and use that ratio to determine your force vs the load weight.  you can do this by making sketches rather than actually setting up real pulleys, ropes and loads.  Good luck.
66380
You have to use the formula -\nm1s1(t3-t1)=m2s2(t2-t3)\n=>191(4.184)(30.5-25)=38s2(192-30.5)\n=>s2=191(4.184)(5.5)/((38)(161.5)\n=>s2=0.716 J/g.k
624334
Example: Find the length of a helix which winds three times around a cylinder of height 12" and radius 3"\n\nAnswer: divide the cylinder into three sections. The helix goes once around a cylinder of height 4" and radius 3"\n\nThe surface of the cylinder can be "peeled" into a rectangle 4" by 2 * pi * 3" = 18.850". The helix is a diagonal of that rectangle. So its length is sqt(4^2 + 18.850^2) = 19.27"\n\nIt goes around the cylinder three times and its total length is therefore 57.81"
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
703613
If they are too strong to pull apart directly, try sliding them.\n\nIf this doesn't work, find an even stronger magnet and put something between it and the other two first.
942101
start by seperating y, u then have y=3x/2+3 then plug in the domain, or the numbers u are givin\nu'll have 0, 1.5, 3, 6, 7.5
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
929849
solve three body problems
35588
They're vectors, so take into account the direction. Thankfully, they're all in the same dimension, so you can make north positive and south negative (or vice versa, if you prefer), then add. -1 + 3 - 6 + 2 = -2, that is, 2 km south.\nIf they were in different directions, you'd do the same thing, adding horizontal and vertical separately, then using the Pythagorean theorem to find the resultant vector.
949810
Use the distance formula:\n\nD = √ [ (y2-y1) ² + (x2-x1) ² ]\n\n1)  D = √ [ (-2+4) ² + (-2+4) ² ] = √(4+4) = √8 = 2√2\n\n2)  D = √ [ (3+1) ² + (-2+7) ² ] = √(16+25) = √41\n\n3)  D = √ [ (4-5) ² + (6+3) ² ] = √(1+81) = √82\n\nNRT