i know there is a density factor involved. how is that used to get the weight per one thouand feet, and the circular mil area of a single strand of aluminum?
206247
0.0145 diameter means Pi x 0.0145 square inches in cross section = 0.0455 square inches. Multiply this by 1000 ft (convert feet to inches by using 12000inches) gives you 546.6 cubic inches for the entire piece. The density of aluminum is\n2.70 grams per cubic centimeter, so you convert 546.6 cubic inches to cubic centimeters by multiplying by 16.387 giving a total of 8,957 grams altogether.
613028
There are too many formulas to calculate that answer. You have to take in account for example, position of object to find potential energy, mass and the weight due to the gravitational force, also Force that is going to be exerted the object and the amount of that energy that will be lost. SO it is a lot to  find that out. But their is not one formula for it. Sorry, But if you do find out Let me know.
334183
your problem is statically determinate. so, use static equations to calculate the tensile force in the cable. maximum sag will occur at the center of the cable where the load should be placed.\n\nyoure asking for deflection and the force. it cannot be. the maximum allowable sag should be used to calculate the tensile force, either, calculate the sag using the allowable tensile force. this can be solve based on the area of the section and the allowable stress.\n\nI hope I was able to answer your question. better if you could send me the figure. my email is exaatto2000@yahoo.com or rbra_engrs@yahoo.com\n\nruel
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Gravity, its weight, friction\nThat's about it.
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sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
391153
Trick question... all things regardless of weight fall at the same speed. You need to know how big the item is first... this is the only way to calculate resistance.
527473
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
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First, I'm assuming that your question is "what do i do to make sure the length and mass per unit length of the wire is constant AS tension changes?" (note change of 'and' to 'as') and the rest of the material is just for explanation.\nThe answer is, you can't keep M/L constant. If your setup actually uses two fixed supports, L will be constant, but tensioning the wire stretches it, removing some M from between the supports. You can measure the amount of stretch by making two benchmarks (maybe small pieces of masking tape) in the unstretched wire and measuring their distance as you change the tension. (Doing the appropriate scaling between the baseline distance and the length of the stretched wire.)\nFor the theory on wire tension and resulting dimension changes, search Poisson's ratio. Basically stretching a length L of wire by epsilon*L (epsilon is a small number) would result in a thinning of radius R of 0.5*epsilon*R if the material were incompressible (thus inexpandable). However all solids including steel expand under tension to some extent, so the Poisson ratio of 0.5 above is more typically 0.25 to 0.35 for metals. See the ref.
147983
I went to Nasa's website, which has an explanation about this. I calculated it out and it turned out to be 145 pounds, rounded to the nearest pound. I myself do not exactly understand the equation fully, but I do understand the concept. Hope this helps!
91691
The first thing that you really need to pay attention to is the dynamics of the wire itself and not the physical size.  Your statement is very bias as I could say that why do fiber optic lines tranmit so quickly and the wire holding up the golden gate bridge is not quicker???\nThe quickest and best answer unless you have degrees in physics is the density of the fibers.  It all comes down to density and a mirid of other factors, but density is the key my friend.\n\nHappy conducting my fiber friend!
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If I'm reading your half pound rule right, the weights (with the metal included) would be:\n1, 3, 9, 27, and 81.\nIn other words, you should put in the following amounts of sand:\n0.5, 2.5, 8.5, 26.5, and 80.5.\n\nHere is the reason for it, though I'm not really sure how to write things like summation and subscripts here.  I will write a subscript in a square bracket.\n\nObviously we can use all possible sums of these weights to balance out our desired weight on the other side of the scale - weights of 1, 3, and 7 give values of 1, 3, 7, 1+3=7, 1+7=8, 1+3+7=11.  But we can also use all possible differences.  For example, if you had an 8 pound weight and a 3 pound weight, you could measure 5 pounds by putting the 3 and 8 pound weights on opposite sides.  Then you could measure some 5 pound object by putting it on the 3 pound side.\n\nLet x[0]...x[n] be the value of n weights (metal included).  In this case n=4 because we go from 0-4.  For simplicity in writing, define y[i] = x[0]+...+x[i] for any i.  The maximum amount we can balance on the scale using the weights x[0]...x[n] is y[n], by putting all weights on one side.  We need to find x[0]...x[n] such that for a given m up to y[n], we can add or subtract members of {x[0]...x[n]} to reach m.\n\nNow we prove this by induction.\nWe propose that x[n+1] = 2*y[n]+1 satisfies our desired property.\nDefine x[0]=1.  Then y[0]=1, so x[1]=2*1+1=3.  This satisfies our property because y[1]=3+1=4.  Clearly 1 and 3 can be weighed.  3-1=2 and 3+1=4, so we can reach all numbers up to 4=y[1].\nNow suppose you have x[0]...x[n] with this property.  So any number up to y[n] can be made by sums and differences of x[0]...x[n].\nNow we include x[n+1]=2*y[n]+1.\nYou can still get from 1 to y[n] by just not using weight #n+1, since the property holds true for x[n].\nBy putting weight #n+1 on the lighter side of any of the y[n] combinations, you can get\nall values from x[n+1] down to x[n+1]-y[n] = 2*y[n]+1-y[n] = y[n]+1.\nBy putting weight #n+1 on the heavier side of any of the y[n] combinations, you can get\nall values from x[n+1] up to x[n+1]+y[n] = x[n+1]+x[n]+x[n-1]+...+x[0] = y[n+1].\nSo with our choice of x[n+1] we can get all values from 1 to y[n+1].\nIn other words, our property holds true for x[n+1].\nBy induction, the property holds true for all x[n].\nUsing the formula to calculate the first 5 values, you get 1, 3, 9, 27, 81 - the powers of 3.
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You could just use the old force = (G*M*m)/r^2\nWhat is missing from the diagram is that the small weights were held by a long thin wire.  The wire formed a weak torsional spring.  If there were some gravitational force, it would cause the weights to twist the wire until the torque of the twisted wire matched the force of gravitation (times the radius).  He could measure the amount of twist by shining a light at a mirror attached to the weights and seeing how far the reflection moved.
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Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
483512
Thermal conductivity: \nCopper: \n9.2*10-2 (kcal/sec)/((m^2)(Cdegree/meter))\nAluminum: \n4.9*10-2 (kcal/sec)/((m^2)(Cdegree/meter))\n\nWe need to assume the rods don't loose any heat to the surroundings. Therefore the kcals conducted through both rods is equal. To calculate the heat conducted per second by the copper, multiply its conductivity, \n\n9.2*10-2 (kcal/sec)/((m^2)(Cdegree/meter)),\ntimes (A*50Cdegree/.88 m).\n\nSet that expression equal to aluminum's conductivity times A*50Cdegree/L. Simplify and solve for L.\n\nEdit: I don't know why the conductivity lines are truncated. Maybe Yahoo is making sure it is compatible with any screen resolution. Anyway, the truncated lines, that I see, should end with \n")(Cdegree/meter))".
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well, the magnetic dipole moment of a loop is given by mu = I*A where I is the current and A is the area of the loop.  so, to maximize the area for a given perimeter, i think you want a circular loop.\n\nhttp://www.faculty.fairfield.edu/jmac/ther/superlatives.htm\n\nas for whether to make a single loop or lots of loops, consider how much area you get for a wire of length L.  the circumference of a circle is C = 2*pi*r, and the area is A = pi*r^2.  solve the first equation for r and plug it into the formula for area, and you get A = C^2/4pi.  here C is simply the length of wire you use.\n\nso, if you make N loops of length L, then you get a total effective area of N*A.  but, if you make a single loop of length N*L, then you get a total area of (N*L)^2/4pi = N^2 * A, which is a factor of N more area than the case with N loops.  \n\nit's because of the r-squared term in the area - as you increase the amount of wire, the area goes up like the square of the length of wire.  whereas if you just add extra loops, it only goes up linearly with the length of wire.\n\ncheers!
8175
The density of hdpe is 0.95-0.97 g/mL.  So do the math: mass (aka weight here) = Volume * Density.  I'm assuming you're capable of that or else you wouldn't be asking about hdpe.
133672
Imagine the stretching of the surface of the drum as the stretching of a spring. use hooke's law to do calculation. u will need to know about the strength of material and etc. very very very complicated