A cylinder 60cm high has a circumferance of 20cm. A string makes exactly 4 complete turns around the cylinder while its two end touch the cylinder's top and botton. How long is the string in cm?
214635
100 cm, its a 3:4:5 right triangle.\n\n[lay the cylinder out flat, if you will.  4 times around is 80 cm, and the width is 60 cm.  So the string has to form the hypotenuse of the 60 cm by 80 cm sided triangle.  60 divided by 20 is 3, 80 divided by 20 is 4, and a right triangle is famously 3:4:5, so 5 times 20 is 100 cm]
505366
Imagine the center of the triangle. Draw a line segment from it to a vertex, and then one from the center to a midpoint of a side. This makes a 30-60-90 triangle, so the line segment is (p/2) sqrt(3). This is a radius of the cylinder, so we multiply it by 2 to get the diameter.\n\nd = p sqrt(3)\n\nor, as the question wants it\np = d/sqrt(3) = d sqrt(3)/3
373124
This is not tricky at all. The volume is 1-1/2 cylinders of "height" 50,000 meters and radius 4 meters.\n\nThe formula for the volume of a cylinder is\n\nV = pi * r^2 * h
536195
Circumference of spool = 2PiR\n\nSince the length of the cable is l, the number of times the cable can be wrapped around the spool = l/(2PiR)\n\nEach "wrap" will cover the spool by 2r.\n\nTherefore the total length that can be covered = 2r*l/(2PiR)\n=rl/(PiR)
328847
Cylinder = PI r^2 x 6 r = 6PI r^3\nBalls = 4/3 x PI r^3 x 3 = 4 PI r^3\nratio = 4 PI r^3 / 6 PI r^3 = 4/6\n\nRatio = 2/3 of the space is used by the balls.
443491
i think the length of a rod with time period t is:\nL=t^2 g sqrt(12) /(8pi^2)\nthis gives legth = 0.211 meter or 21.1 cms\nnot sure about the formula done that long time back  :)
433163
1.  = 1cm\n2.  = 2cm\n3.  = 4cm\n4.  = 7cm\n5.  = 11cm\n6.  = 16cm\n7.  = 22cm\n8.  = 29cm\n\ntotal length = 92cm\neach step adds more to the next length.  \nbetween the first and second step, add 1\nbetween the second and third step, add 2\nbetween the third and fourth step, add 3\nbetween the fourth and fifth step, add 4....\n\nand so on.
497285
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
119919
One full rotation of the tip is:\n\n2 pi r \n2 pi 9.8cm\n\nin 25 min the tip will make 0.42 rotation (25/60)\n\nso 25/60 * 2 pi 9.8 =  25.7
458748
The acceleration of the system can be found using F=ma\nor a=F/m\na=65/(12+24+31)\n=.97 m/s^2\n\nfor the tensions, use the same equation considering the forces on each string.\n\nfor m1 the force is what is pulling the mass so that it accelerates at .97m/s^2\nF=12*.97\n=11.64 N\n\nfor the string between m2 and m3\nF=(12+24)*.97\n=34.92\n\nj
495325
T=2pi*sqrt(L/g) for pedulum with center of mass at the end, so rod length is twice that, so T=2pi*sqrt(L_rod/2/g), the answer is L=g*T^2/2/pi^2=0.243m.\n\nThe length is not 12.2cm because the center of mass of the rod is half distance of the length of the rod
543451
For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
43308
position = (-g * t^2) / 2 + Vo * t + Po\n\nthe two positions are equal to 20m and they are equal to each other.\nVo of the one at the top is 0 and Po is 40.\nVo of the one at the bottom is what we're solving for and Po is 0.\n\nthe equations become:\n\ntop:  (-g * t^2) / 2  + 40 = 20\nbottom: (- g * t^2) / 2 + Vo * t = 20\n\nsolving the top for t gives sqrt(40 / g) which is about 2.02 s.\n\nplugging that into the bottom and solving for Vo gives about 19.8 m /s.
512294
Say the length in centimeters is l\nThen the bredth(b) is 3 less than that, \nb = l - 3\n\nNow the bredth, length and diagonal(d) form a right triangle, \nso we can say d^2 = b^2 + l^2 by pythagorus.\nd^2 = (l-3)^2 + l^2\n15^2 = (l-3)(l-3) + l^2\n225 = l^2 - 6l + 9 + l^2\n216 = 2l^2 - 6l          div by 2\n108 = l^2 - 3l            move over the 108\n0 = l^2 - 3l - 108\nYou could use the quadratic formula, or factor.\n0 = (l-12)(l+9)\nl = 12\n\nb = 12 - 3 = 9\nThe bredth is 9cm and the length is 12cm.\n\nSince the other diagonal is just the diagonal flipped along the center, it is the same length: 15 cm.
154430
It is a problem in long division with a remainder. You need to simply divide the whole integers (everything before the decimal point) and keep the remainder as the inches.  Thus your problem becomes 20 feet and 6 inches, you can check your answer by reversing the problem.  In 20 feet there are 20 times 12 inches, or 240 inches with 6.138 inches left over.  Thus the answer is 20 feet and 6 inches (rounding to whole inches as in your example). \n\n-LOVE YOUR NIEGHBOR AS YOURSELF
17200
find the circumference using C=2(pi)r. r is half of the diameter, so thats 2 times pi times 112.5. this is your distance. speed is distance over time. so now you need to convert minutes to seconds so you can get it in m/s. so you do 10 x 60 and that's 600 seconds. so you just divide the circumference divided by 600 and thats your answer.
442519
This is a simple dimensional question.  There are 2.54 cm in one inch.  For simplicity, you should set up the problem like this:\n\n3.33 cm (1 inch/2.54 cm) = 1.31 inches\n\n(that answer is to the correct number of significant figures, since you have only three in your problem)