you cannot insert the scale into the lake in which the lotus floats. You are not allowed to uproot the plant and use any other devices(& I mean anything). So how do you measure its length ?(There is a mathematical answer to this problem and involves a very simple and common method)
125444
walk into the lake and stand next to the lotus, noting how far up your body the plant reaches from the roots, where your toes are.  get out of the lake, and measure to that point on your body with the provided scale.
610214
there are a few diff ways for this.  the way i do it is with a rotating lazer level. you can rent them at most rental places, another way is with a transit/builders level but if you dont know how...you will be lost cause you wont know how to read it. or the cheap and easy way.   get some clear plastic hose hold the ends up and fill it with water. create a "benchmark". use a post or somthing else you can tie it to and walk away without spilling. on the other end make a leg for you to rest it on roughley as high as the post. now as you move around you can see if it isflat by marking the level in the tube cause water will seek its own level. i prob left out some things in the explanition but im sure you will pick it up.....it was good enough for the romans and it can work for you.
477518
The CG is between the two scales. let us say it is at a distance of x1 from scale 1 and x2 from scale 2. (assume the plank remains horizontal)\n\nThen, x1*Fg1 = x2*Fg2 (the moments must be equal)\n\n>> x1/x2 = Fg2/Fg1\nand x1 + x2 = 2\n>>\nx1 = 2 * (310/(450+310)) ~ 0.816 m\n\nNow tell me where the feet of the woman are.
15212
The question is:  What should I set on the sub-scale of my altimeter so that the instrument would indicate my elevation if I were on the ground at your station?\n\nANSWER:  QNH
652778
The pond is not likely round enough or a pedometer accurate enough for more than a rough estimate.\n\nA better alternative would be to use Google Earth.  It has a measuring utility that lets you define your path and will tell you distances more accurate than the above.\n\nGood luck.
73582
Grab a penny and a dropper. Cant think of whats its called but it is like a n eye dropper. You can use an eye dropper but you'll have to only use that one for ur expiriments will have to have the same conditions. Anyway insert the liquid of choice into the dropper and one at a time slowly begin to drop drops of the liqiud onto the center of the penny. Because of the water surface tension it will for a little dome over the penny. record your results for when the water surface tension breaks. switch between liqiuds. Go on from there. Hope it helps you out
288259
As with all these type sof question the answer is : it depend sentirely on your scale of measurement.\n\nYou are essentially dealing with a fractal relationship. The root sans hairs theroetically has a smooth and hence finite area. The root with hairs has a convoluted area which is potentially infinite.
204068
You could place a vertical rod in the pond (on a stand or stuck in the mud?) and start the experiment when the rod is dry.  After you create the wave and the surface returns to rest, measure the height of the wave on the rod.  This will be the maximum height of the waves at the given point and will depend on the location of the disturbance (radiates outwards like a bullseye?).  Also, there should be nothing near the rod to magnify the wave (including shore effect or irregular bottom?).  Applying chalk to the rod (especially blue chalk?) may make the wetted height more visible and resist drying too fast.  Including a scale on the rod may allow you to read the wetted height above calm surface easier, especially using binoculars.  The trough should approximately equal the wave height.  Good luck.
152103
Make right triangles with the height as one leg, a radius (segment from center of base to one of its vertices) as the other leg, and lateral edge as the hypotenuse.  The radius will equal the length of the base edge as it's a side of an equilateral triangle.  So 10^2 + 15^2 will equal lateral edge squared.\n\nSlant height would be the hypotenuse of a right triangle with legs of the pyramid's height, and the apothem (distance from center of base to midpoint of a side.)  You can get the apothem from Pythagorean theorem from the right triangle formed by half a base edge, a radius, and the apothem.  Then use the apothem in another Pythagorean theorem to get slant height.\n\nLateral area = 6 times the area of any one triangle.  Each triangle's area is 1/2 times 10 times slant height.\n\nVolume = 1/3 (area of base)(height).  The area of the base is 1/2 (apothem)(perimeter).\n\nThose are the steps.  Hope you can figure them out from this typing!
366295
People usually don't do the Sun to scale when they do their model. They do it smaller. And they'll use distances between planets that are not to scale for the real distances from the sun.\n\nThat's the straight answer. Use artistic license, since using scale is not practical.
277507
I think let the Drosera grow naturally,with normal water.Please don't interfere with it's life too much.
536971
The weights of the liquids inside 2 branches must be equal, therefore:\n\na) m_water + m_oil = m_mercury (the amount of mercury that raised in the other branch to keep the balance)\n\nm_water = D_water * V_water = D_water * h_water * Area\nm_oil = D_oil * V_oil = D_oil * h_oil * Area\nm_mercury = D_mercury * V_mercury = D_mercury * h_mercury * Area\n\nSubstitute this in the balance of weights above, we have\nD_water * h_water * Area + D_oil * h_oil * Area  = D_mercury * h_mercury * Area\n<-> h_mercury = (D_water * h_water + D_oil * h_oil) / D_mercury\n\nb) To bring mercury's surface two the same level again, the weight of the to-be-added water must be the same with the weight of both oil and water in the other branch.\n\nYou can do the same as above to find:\nh_water2 = (D_water * h_water + D_oil * h_oil) / D_water
127482
This is a good question that made me think for a while.  First of all, you need to know how big Earth is.  Eratosthenes did that in ancient times by measuring the altitude of the sun at two different places on the Earth at the same time on the same day (presumably he had a friend helping him).  I've linked to a site detailing this measurement.\n\nOnce you know that, just wait for a lunar eclipse.  When the Earth is partially obstructing the moon, you will be able to see an arc of the Earth at the same time as an arc of the moon.  Then you can get a relative difference in curvatures between the Earth and the Moon (maybe by using coins of different sizes, for instance).  So once you know the ratio of the Moon's circumference to the Earth's circumference, just multiply by the Earth's cirumference to get the answer.
289681
its very easy some amount of sand & compost (made from animal waste) thoroghly spoiled. mix them in equal quantity and add some water in it to keep it wet. now just deep a part of cutting in it where you want roots to come. also a solution come here in india's market if this not work.
323445
yup your answers are correct.\n\nhowever your units are not, length and width are measured in yards not sq yards.
16533
Physics, major science dealing with the fundamental constituents of the universe, the forces they exert on one another, and the effects of these forces. \n\nMeasurement, finding the size or amount of some quantity, and expressing it as a number of defined units.All measurement is based on a comparison. In general, the thing to be measured is compared with something that has already been calibrated, or measured against a known reference. However, a measurement is not complete unless it includes a statement of its uncertainty, or the confidence one can have in it.\nUnits of Weight and Measurement\nImperial System\nAMOUNT NAME \nAngular or Circular Measure \n60 seconds = 1 minute (min) \n60 minutes = 1 degree \n30 degrees = 1 zodiac sign \n57.2958 degrees = 1 radian \n90 degrees = 1 quadrant or right angle \n360 degrees = 1 circle \nCubic Measure (Volume) \n1,728 cubic inches (cu in) = 1 cubic foot (cu ft) \n27 cubic feet = 1 cubic yard (cu yd) \n231 cubic inches = 1 US gallon (gal) \n277.27 cubic inches = 1 British imperial gallon (i gal) \n2,150.42 cubic inches = 1 US bushel (bu) \n2,219.36 cubic inches = 1 British imperial bushel (i bu) \nDry Measure (Capacity) \n2 pints = 1 quart \n8 quarts = 1 peck (pk) \n4 pecks = 1 bushel \nLinear Measure (Length) \n1,000 mils = 1 inch (in) \n12 inches = 1 foot (ft) \n3 feet = 1 yard (yd) \n5.5 yards = 1 rod (rd) \n40 rods = 1 furlong (fur) \n8 furlongs = 1 mile (mi) \n5,280 feet = 1 mile \n3 miles = 1 league (l) \nLiquid Measure (Capacity) \n4 fluid ounces (fl oz) = 1 gill (gi) \n4 gills = 1 pint (pt) \n2 pints = 1 quart (qt) \n4 quarts = 1 gallon (gal) \nMariners’ Measure \n6 feet = 1 fathom (fath) \n120 fathoms = 1 cable’s length (US Navy) \n1,852 metres = 1 nautical mile (n mi) \n1 nautical mile per hour (n mph) = 1 knot (k) \nPaper Measure \n25 sheets = 1 quire (qr) \n20 quires = 1 standard ream (rm) \n516 sheets = 1 printers’ ream \n2 reams = 1 bundle (bdl) \n4 bundles = 1 case (c) \nPrinters’ Measure (Typography) \n0.013837 inches = 1 point \n12 points = 1 pica \nSquare Measure (Area) \n144 square inches (sq in) = 1 square foot (sq ft) \n9 square feet = 1 square yard (sq yd) \n30.25 square yards = 1 square rod (sq rd) \n160 square rods = 1 acre \n640 acres = 1 square mile (sq mi) \nSurveyors’ Measure \n7.92 inches = 1 link (li) \n66 feet = 1 chain \n80 chains = 1 mile \nWood Measure \n144 cubic inches (1' x 1' x 1'') = 1 board foot (bd ft) \n16 cubic feet (4' x 4' x 1') = 1 cord foot (cd ft) \n8 cord feet = 1 cord (cd) \nWEIGHTS \nWeight (Apothecaries’) \n20 grains = 1 scruple (s ap) \n3 scruples = 1 dram (dr ap) \n8 drams = 1 ounce (oz ap) \n12 ounces = 1 pound (lb ap) \nWeight (Avoirdupois) \n27.3438 grains = 1 dram (dr) \n16 drams = 1 ounce (oz) \n16 ounces = 1 pound (lb) \n14 pounds = 1 stone \n100 pounds = 1 hundredweight (cwt) \n8 stones = 1 long hundredweight \n(l cwt) \n2,000 pounds = 1 short ton (ST) \n2,240 pounds = 1 long ton (LT) or 20 (l cwt) \nWeight (Troy) \n24 grains = 1 pennyweight (dwt) \n20 pennyweights = 1 ounce (oz t) \n12 ounces = 1 pound (lb t) \n\nMiscellaneous Units\nWEIGHT/MEASURE UNITS \nAbsolute zero = -273.15o C (-459.67o F) \n1 atmosphere = 760 mm of mercury \n1 cup = 10 fluid ounces \n1 carat (international) = 200 milligrams (gem weight) \n1 megaton = 1,000,000 tons \n3 teaspoons = 1 tablespoon \n1 angstrom unit = 0.1 nanometre \n1 bolt of cloth = 40 yards \n1 gross = 12 dozen, or 144 \n1 light-year = 9.46 x 1012 km \n(5.88 x 1012 mi) \n1 tablespoon = 1/2 fluid ounce
274579
i haven't done this in a while.\n\nThe volume of the reservoir is equal to (1/2)*(height)*("base")*length.  the reservoir is a rectangular prism with the "base" being the distance across the surface of the water.  We are given the shape of the full volume so we can scale down to the level of the water to find what the volume in the reservoir at its current level.\n\nif you think about the cross-section of the volume, and use half of that forming a right triangle, we can determine the ratio of the surface (the "base") to the height (water depth).  In this case, it is (40/2)/(20*sqrt(3)) or 1/sqrt(3).  so ("base"/2)/(height) = 1/sqrt(3)  and then "base" = (2*height)/sqrt(3).\n\nsubstitute that into the volume equation so that\nvolume = [(height^2)*length]/sqrt(3) or literally,\nV = [(h^2)*L]/sqrt(3).  differentiate with respect to time,\n(dV/dt) = [2*h*(dh/dt)*L]/sqrt(3) and substitute the numbers in:\ndV/dt = 200 - 50 = 150\nh = 10*sqrt(3)\nL = 450\nand solve for dh/dt which is the rate of change of the water level in yds/hour.  these are odd units, you might want to change to inches/hour.\n\ni did it this way and got 1/60 yds/hr or 0.6 inches/hr.\n\nbut again it has been a long time....