If you fold a piece of paper .1mm thick 40 times, how do you figure its thickness?
155820
Each time you fold it, it doubles in thickness and halves it surface area.\n\n0.1mm x 2 ^ 40 = 109,951.1627776 km\n\n(68,320 miles)\n\nThat means that the piece of paper would be so thick that it would reach to the Ionosphere (well past the ozone layer!)\n\nAlthough the thickness of the paper would be huge, the surface would be tiny. Say you had a large sheet of paper\n1m x 1m (3'4" x 3'4")\n\n1(m^2) / 2 ^ 40 = (m^2) = 0.90949 (μm)^2\n\n(0.0000000014097 square inch)\n\nSo basically your talking about a large piece of paper that now has such a small surface as to become invisible and so thick that it may get hit by international aircraft!\n\nThat is ridiculous and impossible! Incidentally, no paper can be folded more than 8 times.
709877
1 inch = 25.4 millimeters.\n\nHere's a calculator\nhttp://education.yahoo.com/reference/weights_and_measures/index?p=length
1058416
It is on old trick to bet someone that they cannot fold a piece of paper 9 times.  Even after 5 or 6 foldings it's not creased anymore, but rather bent.  \n\nIf you fold a piece of paper 9 times, each fold doubles the number of sheets, so you have 2 to the 9th power of sheets, or 512 sheets of paper in thickness (just over 1 ream).  Ever try to fold a ream of printer paper?  \n\nHere's a mind blower for you:  If you could cut (or fold) a sheet of paper 100 times, how thick would it be? \n\nThe answer:  Assuming that each sheet of paper is 100 microns (1/10 of a mm) thick, you would have 1 Googol sheets of paper (10 to the 100 power), and the stack would be about 13.6 billion light years tall, which, incidentally, is more than the estimated diameter of the known universe (about 12.8 billion light-years, according to astronomers).  \n\nNever underestimate powers of 2 ;-)
1133339
1cm/15m=Xcm/40m\n\nX=40/15=8/3cm\n\nI hope that does the trick
749682
1. The initial dimensions of the cardboard before cutting are 15 x 24. We cut out equal squares of side x, leaving \n\nl = 24 - 2x\nw = 15 - 2x\nh = x\n\nthe volume is\n\nV = (24 - 2x)(15 - 2x)x\n = 4x^3 - 78x^2 + 360x\n \nLeave us take the first derivative of this, to get the maximum:\n\n0 = 12x^2 - 156x + 360\n = x^2 - 13x + 30\n = (x - 10)(x - 3)\n\nThe solution x = 10 is not reasonable, since it will leave us with negative width. Using x = 3, we get\n\nl = 18\nw = 9\nh = 3\nV = 486\n\n2. This is a downward-opening parabola that has the y-axis as its axis of symmetry and a maximum at (0, 8). HayHrbr's analysis is correct, and the first derivative of the area gives us\n\n16 = 12x^2\nx^2 = 4/3\nx = 2 / sqrt(3)\n = 2sqrt(3) / 3
163967
It's not technically 7 or 8. It's just that for ordinary paper that you have hanging around the house -- looseleaf, newspaper, wrapping paper -- all of these are approximately the same order of magnitude for their width/size in respect to thickness. \n\nThe first fold doubles the thickness and halves the width/length. The second doubles again, and so on. Essentially, if the original width/length is W and thickness T, for each fold N, the new width is W*2^(-N) and new thickness is T*2^N. For a piece of looseleaf, whose original length is about 270 mm and thickness is 0.1 mm, the 7th fold causes the thickness to surpass length. For a post-it note, the fold that does it in is theoretically the 5th or 6th. The site below talks about a bright girl who decided to roll out 4000 feet of toilet paper in a mall. Toilet paper is about 0.05 mm thick so theoretically, the 13th fold should be tough. She got 12 out of it.
325473
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)
191768
r u sure its not 40 m less, if it is here:\nLet w=width\n    w+40=length\nw(w+40)=21^2\nw^2+40w=441\nw^2+40w-441=0\n(w+49)(w-9)=0\nw=-49 or w=9\nthe width is 9 m and the length is 13 m
526883
set up a needle that is bent at the end. put it on the ground. then get a giant roll of paper and prop the bent end of the needle on to it. then make a machine that will turn the paper. dip the bent end in ink, and you have a siesmometer
1095598
For part 1, if the cell's diameter was 1/15 of the total field, and the field was 450 microns across, then the cell's actual diameter was 450 microns/15 = 30 microns = 3 * 10^-5 meters.\n\nThe diameter of this cell in the drawing is 0.5 cm = 5*10^-3 meters.  This drawing therefore has a magnification of (5*10^-3)/(3*10^-5) = 166.6666666...times.\n\nFor part 2, The magnification is given by (size in image)/(actual size), so we have that 45000 = (1.5 mm)/a, where "a" is the actual thickness.  Solving for a, we get a = 1.5mm/45000 = 3.333*10^-8 meters = 33.3 nanometers = 0.0333 microns.\n\n-----------------------\n\nAs for the question posed in the "additional details", when the magnification of a microscope is increased, the diameter of the field being imaged decreases (choice "A").
458726
you can fold any peice of paper up to 7 times. wont go anymore than that.   Just a bit more dumb trivia i know
272896
Two 20.Rs notes, Two 2.Rs notes , and Six 1.Re note.That is 20x2 = 40 + 2x2= 44 + 1x6 = 50.\n\nWas that the answer you were looking for?
409297
ok here is the thing\nif you are trying to calculate the thickness of a piece of paper you will need some equpment. Even though there is a lot of room for human error you still can get a fair estimate. \n\nwhat you need is 2 extrememly smooth glass plates and a lineat light source say a laser pointer. \n\nput one glass plate on a table place your paper on the side of the glass and put the other glass plate on the top of the two so that you will get a wedge between the two glasses. \n\nnow point the laser pointer perpendicular to the bottom glass plate and place a screen for reflections. you will get multiple reflections. (btw you need to know the frequency of the light emitted from the pointer.)\n\nthis is the set up\nfor calculations do a little search on air wedge interference \n\ni dont think you will need refractive indexes since you are using the same glass\nwith a little algebra you can figure it out
13634
It gets too thick to be folded I guess, try it using tracing paper it should work
234958
1 mil is 1/1000 of an inch.\nSo if something had a thickness of 1 mill, if you had 1000 of them stacked together, it would have a thickness of 1 inch.
783277
let the height be x\nthe cross section will be (40-2x)*x\ndA/dx=40-4x\nsetting this to 0\nx=10\nso the length of the sides will be 10 cm each for max area of x section
968499
1) If they shared the delivery equally, then they would each deliver p number of papers but Erin delivered 20 of Jennifer's, so \nGilbert = p\nErin = p + 20\nJennifer = p - 20\n\nTotal = p + (p + 20) + (p - 20) ---- (1)\n\nSince erin's delivery is 1/2 of the total papers, we have that \n\np + 20 = [p + (p + 20) + (p - 20)]/2 ----- (2)\n\nFrom (2) we have that \n2(p + 20) = p + (p + 20) + (p - 20)\n2p + 40 = p + p + 20 + p - 20\n\nCollect like terms\n40 + 20 - 20 = p + p + p - 2p\n40 = 3p - 2p\n40 = p\np = 40\n\nSo Gilbert delivered 40 papers\n\n\n2) Considering the last sentence - one-fourth of those remaining left, leaving only 12 people in the audience. Since 1/4 left leaving 12, it means that 12 is 3/4 of the audience\n\nIf 3/4 = 12\n4/4 = (4/4 X 12)/(3/4)\n= 4/4 X 12 X 4/3\n= 16\n\nRemember that 1/3 left remaining 16, so 16 was the 2/3 of the audience.\n\nIf 2/3 = 26\n3/3 = (3/3 X 16)/(2/3)\n= 3/3 X 16 X 3/2\n= 24 \n\nAlso, remember that this fugure was just 1/2 of the original audience\nIf 1/2 = 24\n2/2 = (2/2 X 24)/1/2\n= 2/2 X 24 X 2/1\n= 48 which is the total audience\n\nCheck ...... working backwards\nAt ther beginning, there were 48 people in the audience;\nAfter a few minutes, 1/2 left remaining 24\n\nFive minutes later 1/3 of the remaining people left;\n1/3 of 24 is 8, so if 8 left, we have 24 - 8 = 16 people left.\n\nTen minutes later 1/4 of the remaining people left;\n1/4 of 16 is 4, so if 4 left, we have 16 - 4 = 12 people left!