How can I calculate the critical Reynolds number knowing the diameter of a sphere?
965871
If the sphere is not inside a relatively narrow pipe, the number is about 3.85 * 10^5. There is no formula for it, it is defined experimentally - see slide 23 in [1]. A short description of the phenomenon itself is, for example, in [2]:\n------------------------\nIt is known that a sphere in a steady flow will exhibit differing drag coefficient values depending on whether the boundary layer on the sphere surface is laminar or turbulent. The free-stream velocity at which the transition from laminar to turbulent flow occurs depends on the diameter of the sphere and the turbulence level of the free-stream flow. It is also known that in an atmospheric free-stream the Reynolds number for transition is approximately 385,000. At this critical Reynolds number, the drag coefficient of a sphere drops from a value > 0.4 to a value < 0.2. By measuring the Reynolds number at which this drop occurs in a wind tunnel for different size spheres, one may determine the turbulence level at different velocities. \n------------------------\n\nThis should be enough info for you to make a deeper search on the Internet if needed.
135768
Area = Pi times the radius squared (A=pr2)\nIf you need to know how to find the radius, go here:\n\nhttp://math.about.com/library/blcircle.htm
341043
Previous answerer had it correct.  Falling-ball viscometers are based on Stokes' Law.  See sources below for a complete description.
42062
Surface Area of a Sphere = 4 pi r^2\n\nCompute it for 95mm and 130mm and take the difference.
408251
Measure the diameter and multiply by pi (3.1415)
67284
well the formula is the square root of 3kT/m  where k is the boltzman constant(1.38066 x 10^-23 J/K) T is in Kelvin so you would have to convert(just add 273 to the temp in Celcius) so your T is 290K and m is the mass. the mass can be found using the density of water(1000 kg/m3) the formula for density is d=m/V and V can be found by using the volume of a sphere, 4/3(pi)r^3 oh, and don't forget to convert the nanometers into meters, then cut the diameter in half for radius.\nusing this information to get you started, you should be able to do it yourself. so you should try it first before reading further :) im sure you were familiar with all of those formulas, you just need to remember to think of things all at once so you can put them together.\n\nok, so we start with the rms velocity formula. first problem is that we need mass. ok, lets solve for mass using density. oh right, we need volume, ok, lets solve for volume then:\nwe need to make the diameter into meters, so thats 1.0 x 10^-8 meters. now we cut this in half to get the radius(r) so that's 5.0 x 10^-9.\nV=(4/3)(pi)r^3\nV=(4/3)(3.14159)(5 x 10^-9)^3\nV=5.236 x 10^-25 m^3\nnow we have volume, so using the density of water(1000 kg/m3), we can solve for mass:\nd=m/V\n1000=m/(5.236 x 10^-25)\nm=5.236 x 10^-22 kg\nok, now we have mass, so we can use our original rms velocity formula and solve for velocity. *sigh* finally.\n\nv(rms)= sqrt[3kT/m]\nv(rms)= sqrt[3(1.38066 x 10^-23)(290)/(5.236 x 10^-22)]\nv(rms)= 22.94 m/s\n\ni would check my math if i were you though, i tend to make stupid calculator errors. the concepts are there though :p
802377
Assume that the sphere is centered at (0,0,0)\nStart out by slicing the sphere through its center and looking at the cross section of the sphere and the cylinder. You will have a rectangle inscribed inside a circle.\n\n\nDraw a line from the center of the rectangle to the corner of the rectangle that is in the first quadrant. It will have length r since it is a radius of the sphere.  Let t be the angle the radius makes with the positive x axis. \n\nThe height of the cylinder is   r*sin(t).\nThe radius of the cylinder is r*cos(t)\n\nThe surface area is 2*pi*r^2cos^2(t) + 2*pi*rcos(t)rsin(t)\n\nTake the derivative and find the max.
1001159
it is 4/3*pi*r^3\n\ngo to this website for an explanation\n\nhttp://www.qc.edu.hk/math/Junior%20Secondary/Sphere.htm
525014
Let's make this very simple.\n\nTo measure the volume of a cylinder, you find the diameter of the interior portion of the cylinder.  Take that number and divide by two...that's your radius (r).  Now measure the height of the interior of your cylinder...that's your height (h).\n\nNow, take the radius (r) and the height (h) and plug them into the formula below:\n\nV = π*(r)*(r)*(h)\n\n(π equals roughly 3.1516)\n\nNow if your answer is in cubic inches, multiply V by .0036 for British gallons or .00433 for US gallons (yes, they are different).  If V is in cubic centimeters, multiply by .00022 (British) or .00026 (US).\n\nI don't know how to make it any clearer...good luck.
894074
Assuming the drop of water is a sphere or close to one...\n\nvolume of a Sphere = (4/3) * pi * radius^3\n\nTo find the volume of a drop, measure out a small volume of water into a graduated cylinder.\nUse a dropper to take drops and deposit them into another container.  Count the number of drops needed to empty the measured water.\nvolume of drop = measured volume / number of drops
955775
Type "linear coefficient of thermal expansion" and "volume coefficient of thermal expansion" into you search engine.\nFor spheres, you probably want to use volume.\n\nYou should find formulas that relate change in length (or volume) and deltaTemperature.\n\nPlug the two temperatures and the coefficient into the formula and you will have the change in size.
981062
The surface area of a sphere of radius 'r' is given by:\n\nsurface area = \n 4*pi*r^2\n \n\nNOTE: The value of p can never be known exactly, so surface areas of spheres cannot be calculated exactly. Common approximations for p are: 3.14, and 22/7. \n\nExample 1  if 'r' = 5 for a given sphere, and p = 3.14, then the surface area of the sphere is:\n\nsurface area = 4*pi*r^2\n\n         = 4 × 3.14 × 25\n\n         = 314
1140416
you need to employ th ebournoulli equation and the fanninf (or darcy ) friction factor equation ... \n\nThe Darcy-Weisbach Equation for pressure loss due to friction (in the SI metric system) is:\n\nΔP = 4f(L/D)(ρ v2)/(2 gc)\n\nwhere: \nΔP = pressure loss in Pascals \nf = Fanning friction factor (dimensionless) from Moody's diagram \nL = pipe length in meters \nD = pipe diameter in meters \nρ = fluid density in kilograms per cubic meter \nv = fluid velocity in meters per second \ngc = gravitational conversion constant = 1 (kg-m)/(N-s2) in the SI metric system\n\nAnd for head loss due to friction, the equation is:\n\nΔh = f(L/D)(v2)/(2 g)\n\nwhere: \nΔh = head loss in meters \nf = friction factor (dimensionless) \nL = pipe length in meters \nD = pipe diameter in meters \nv = fluid velocity in meters per second \ng = local gravitational acceleration of 9.807 m/s2
757325
Use (P1V1 / T1 = P2V2/T2)  \n\nP is constant ..First determine V1 by substituting 1 cm in volume of sphere equation.  So you will get V1. This is V1 when T1 is 280 K.\n\nThen use new temp 361K and find out what would be V2  using above equation.  Once you get new V2, plug it in the volume of sphere equation and get new radius.  \n\n\nGood luck.
653555
Not considering the spheric harmonic of high orders I would have to tell you that my physics spirit tell the surface area  is what counts and there you could have k0.2Pi.R=2Pi <=>k0=1/R\nk1=1/(2R), k2=1/(distance of the 3 points more away from each other)........................
791771
Equatorial diameter = 12,756.2 km\nPolar diameter = 12,713.6 km\nVolumetric mean diameter = 12,742.0 km\n\n(BTW there is no such thing as centrifugal force, it's an engineers' myth; there is only inertia.)  :)
125839
I'm not sure. But the volume of a sphere is (4/3)(pi)(r)^(3). \n2) You know that the diameter is double the radius. So your radius is 1.\n3) Then, your volume is (4/3)(pi). \n4)Density=(mass/Volume)\n5)Density of water is 1.00 mL=1.00g/m^3\n6)MASS=(Density)(Volume)=((1.0g/m^3)(4/3)pi). Note that cubic meters cancel. The mass is (4/3)pi grams.\n\nHope I helped.