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Electrical current alone has no heating effect. Current through a device, with a voltage across the device will have a heating effect. The equation for calculating it is power = voltage x current, where power is proportional to the heating effect If the AC supply is measured as an RMS voltage and an RMS current and the device is resistive, then the heating effect will be identical to the same values with a DC supply. RMS means the "average" voltage or current of an AC supply whereas the peak AC voltage refers to the highest voltage that is reached on each cycle. However, if the device is not just resistive but is inductive, the heating effect will be lower with an AC supply than with a DC supply. By inductive, we mean that the device has a coil or capacitor, for example, in the circuit. The reasons why are outside the scope of this answer but are explained in many electronics text books, or look up "power factor" on Google
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How do you mulitiply fractions?
a/b*c/d = ac/bd E.g. 2/3*3/2 = 2*3/3*2 = 6/6 = 1.
How do you do linear regression in Casio fx-991ES PLUS?
go to stat mode then then select (A+BX) mode and enter the data and press AC on cal. then shift+1 and go to the stat and select REG and there you can see options like A,B and r u can select any of these to get what u need .if you want the answer for r select that option. thnx.
Volume of a truncated rectangular based pyramid?
(1)Best formula to use is as follows -V = h/3(Areatop + √(Areatop*Areabottom) + Areabottom)(2)To find (h) using a tape measure -Areatop => bdAreabase => acLateral edge remaining => e (from top corner to base corner)k = 1 - √(bd/ac)H= √([e/k]² - [a/2]² - [c/2]²)h = HkV = H/3*(ac-bd+bd*k)(3)Lets say Top is a rectangle with sides b & dand bottom is a rectangle with sides a & c respectively.Let height be hin that case the volume of Truncated Pyramid with rectangular base will be -V = 1/3((a²c-b²d)/(a-b))hBUT BE CAREFUL - a,b,c,d are not all independent variables (one depends on the others) so this answer is misleading!!!Proof -Suppose the height of Full Pyramid is HFrom parallel line property(H-h)/H = b/aRearrangingH = ah/(a-b) --------------------(1)AlsoSince V=1/3 Base area X HeightVolume of full pyramid = 1/3 X ac X HVolume of removed Pyramid = 1/3 X bd X (H-h)So volume of truncated part V = 1/3(acH-bd(H-h))=1/3((ac-bd)H + bdh)From (1)V = 1/3((ac-bd)ah/(a-b) + bdh)reducing and rearranging we getV = 1/3((a²c-b²d)/(a-b))h(4)In case the truncated solid forms a prism instead, we have following formula -V = ( h/6)(ad + bc + 2ac + 2bd)Proof -Fig(1)Fig(2)Lets divide the fig(1) into four different shapes as shown in fig(2)VA = Volume of cuboid = bdhVB = Volume of prism after joining both Bs= ½ X base X height X width = ½ (a-b) (d)(h)VC = Volume of prism after joining both Cs = ½ X base X height X width = ½ (c-d) (b)(h)VD = Volume of rectangular pyramid after joining all Ds = 1/3 X base area X height =1/3 (a-b) (c-d) hThen V = VA + VB + VC + VDOr, V = bdh + 1/2(a-b)dh +1/2(c-d)bh + 1/3(a-b)(c-d)hArranging and simplifying we get -V = ( h/6)(ad + bc + 2ac + 2bd)
What does a 6 foot tall 210 pound man look like?
I myself stand 6' and weigh 210 pounds. I am not an extremely muscular guy but I do have some strength to me but I also do not consider myself as being fat. However I do feel I could live with losing a couple pounds around my waist. http://a667.ac-images.myspacecdn.com/images01/128/l_01e5ea43312d41ccc59b5179cac83012.jpg This is a link to one of the only pictures that shows my upper body, although not a very good comparison picture being as my two friends are roughly the same size as me. I am on the right hand side.
