a loop of area .250 sq. meters is in uniform .020 0-T magnetic field.If the flux through the loop is 3.83x1?
359026
Post in the physics section, somebody will know this.
801553
lol.\nfind formula for area of spiral.  subtract smaller from bigger.  sry no more help.
309325
Since m*C*dT=Q;\n\nHeat capacity=135000 J/(5100 g*(31.5K - 18K))= 1.96 J/gK\n\n(K for Kelvin)
954796
51kg x 25 m = energy used\n\ndivide the energy used by 0.30 to get the second answer
265928
1.27 X 1022 KG (0.0021 Earths)
1146296
The magnetic flux:\nit is the product of area & magneric flux component \nThe magnetic flux density:\nit is numerically measured by the number of magnetic flux lines passing perpendicular to a unit area surrounding that point & it is measured in Tesla
551690
T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area
952353
The curved surface area is 2*pi*r*h\n\nor 2*3.142*27*12 sq cms\n\nand 2*3.142*17*35 sq mms\n\nyou can do the rest
990812
a.flux=integral E.dA=Q/e0\n=3.60x10^-6/8.85x10^-12\n=.406x10^6Nm^2/C\n\nb.780=Q/8.85x10^-12\nQ=780x8.85x10^-12\nQ=6903x10^-12C
814915
Area = ½ ab SinC\n Area = ½(55)(29) Sin 61º\n Area = ½(1595) Sin 61º\n Area = 797∙5 Sin 61º\n Area = 797∙5(0∙874 619 707)\n Area = 697∙509 2165\n Area ≈ 697∙51 square units.
496124
Calculate the electricity.
104622
I am trying to figure how large a .745" round wire can be shaped into a rectangle shape wire. \n\nAssuming the wire is malleable, find the area of a circle with diameter .745 or radius = .3725\n\n(.3725)squared times pi = .4359156156 sq in.\n\nFor a square to have this area, each side would be the square root of that number which is .6602390595.\n\nFor a rectangle that is not a square, pick any 2 numbers whose product is .4359156156.  Choose one dimension and divide by it to get the other.
819160
em energy radiated is proportional to T^4, so it would be : ((273.15+9627)/ (273.15+5227))^4 =10.5
710873
The area of the sheet is 250*200=50000 cm^2, \nand area of the hole is 0.6*50000 = 30000 cm^2. \nArea of the hole is \n=(250-2*x)*(200-2*x) \n=50000-500*X-400*X+4*X^2\n=4*X^2-900*X+50000.\nArea of the hole is= 30000 cm^2. \nEQUATING above two, we have\n4*X^2-900*X+50000=30000\n4*X^2-900*X+20000=0\nX^2-225*X+12500=0\nx=-b+/-sqrt(b^2-4*a*c)/2*a\nx=225+/-sqrt(225^2-4*1*12500)/2*1\nx=225+/-sqrt(50625-50000)/2\nx=225+/-sqrt(625)/2\nx=225+/-(25)/2\nx=(225+25)/2 or x=225-25/2\nx=125 or x=100
917169
L = μ(0) μ(r) N^2 A/l\n\nwhere\n \nμ(0) is the permeability of free space (4π × 10-7 henrys per metre) \nμ(r) is the relative permeability of the core (dimensionless) \nN is the number of turns. \nA is the cross sectional area of the coil in square metres. \nl is the length of the coil in metres\n\nHere we assume that μ(r) = 1, since the solenoid doesn't enclose anything like a metal core. Then\n\nL = [μ(0) N^2 (πR^2)] / l, where R is the radius of the solenoid\n\n= [(4π × 10-^7)(2920)^2(π(0.005)^2)] / (0.1)\n\n= 8.44 × 10-^3 henries
303161
The permeability of space is given the symbol of the greek letter mu.\nThe permeability of free space (vaccuum) has a value of \n4 * pi * 10 ^(-7)     H*m^(-1)\n\nThe permeability of free space is, AND I EMPHASISE THAT THIS IS ONLY A CONCEPTUAL DESCRIPTION, how hard it is to set up a magnetic field in whatever medium we are considering.\n\nAll of magnetics depend on the permeability of free space - it is used in almost every calculation of magnetic field strength.\n\nTry searching wikipedia for "Ampere's Law"\nOr Maxwell's Correction to Ampere's Law (Displacement current)\nThe latter is the most fundamental equation for how currents and varying E fields set up magnetic fields, and, as you will see, the equation has mu in it!!\n\n:)
491209
175.76 Amperes...\nusing m=z*i*t\nz=63.546/96500\nt=24*60*60\nm=10*1000