What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters?
447744
Linear expansivity for aluminum is 23x10^(-6) /K. \n\nIncrease in length is 1.2 x 23x10^(-6) x 50 = 0.0138 m.\n\nTherefore the length is 1.2138 m.
860146
The first value represents the (average) Minimun length recorded, the second the Average and the last the Maximum measured values. \nLength............141,5 mm..........161,5 mm..........181,1 mm. \nCircumference..117,5 mm.........127, 3 mm.........137,8 mm.\nKensey data are quite old. These statistic is the last one.
437503
Q = m.s.ΔT\nHeat loss by the copper = heat gained by the metal.\nCopper (heat loss) = 200.0g x 0.38J/g.ºC x (195.5ºC-228.4ºC) = -2500.4J\ndivide the heat by the mass and ΔT of the metal to get the specific heat of the metal:\n2500.4J/(65.0g x (195.5ºC-18.3ºC)) = 0.2171 = 0.21 J/g.ºC (correct SF)
1045656
An electron passing through a potential difference of 150 kV picks up 150keV of energy.  So:\n150keV=h*c/lambda\nlambda =h*c/150keV.\nh=6.626*10^-34 J*sec\nc=3*10^8m/sec\n1eV=1.6*10^-19 J\n\nSo, lambda=((6.626*10^-34)(3*10^8))/((150000)(1.6*10^-19))\n\nlambda=8.28*10^-12 m, or 8.28 nm
617124
A rod is 16 1/2 feet so 80 rods would be 1320 feet or a quarter mile.
1017844
You need the thickness, or the "height" of the aluminum foil.  Your density equation states that aluminum is 2.70 g/cm^3.  You have all the information except the height.  So plug in the numbers:\n\n2.70 g/cm^3 = 2.04g all over 11.5 cm X 14.0 cm X h\n= 2.04 g / 161 X h cm\n2.70 h= 2.04 / 161\n2.70 h = 0.0127\nh = 0.0047 cm\n\nReplace answer for h and you'll find the answer is correct.
775136
when ever two dissimilar metals are joined  they produce a specific millivoltage at each degree of temperature\nif you graft the changes you will find a bell curve \nunique to each type \nthis is how you determine which is proper one for your use\ntype K alumel and cromel  is used at medium temperature ranges 90 to 1600 degrees F
773883
Stop for a minute... Look over your question. Note 2 things\na. 77.2 AU\nb. An icy object\n\nNow start over slowly and find your mistake...YOU CAN DO THIS!!!
783249
Change it to 1.27 × 10^4\n\nYour value for Specific Heat of Zn is only 3 sigfigs, thus, your answer can only be up to 3 sigfigs.
943189
Mmmmm, methinks this is a trick question but I will guess at minus 2 degrees...;
1031853
The radiation length for Al is 24 cm, for Pb 0.56 cm. To reduce the beam to 10% we need 2.3 radiation lengths, 55 cm of Al or 1.3 cm of Pb. \n\nNo idea how your  attenuation coefficients relate to my radiation lengths. A radiation length is the length of material needed to attenuate the beam by 1/e, that is to 0.368 of the original intensity.\n\nGood luck!
886896
0.696850393"\n1"=25.4mm=2.54cm
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
926295
Convert all this to MKS units and find the cross-section of that wire.\n\nE (stainless-steel) is given to be 18 x 10^10 N/m^2.\nCross-section is from pi(r^2).  This must be in m^2, not mm^2.\n3.14159(0.125*10^-3 m)^2 = 4.909 x 10^-8 m^2 cross-section.\n\nYoung's modulus (E) is given as,\n\nE = (delta F / cross-section) / (delta length / original length)\n\nDelta F is the tension in the wire, thats the first part of solving this problem.  So rearrange the equation for delta F, renaming delta F as "tension" and substitute the value given in place of E:\n\nTension = (18 x 10^10 N/m^2) * (8 x 10^-5 m / 0.028 m) * (4.909 x 10^-8 m^2)\n\nTension in the wire is therefore 25 N.\n\nThe magnitude of force on the tooth is 25 sin 30 deg or 12.5 N\n\nWarning!:  While the reasoning is sound, double-check all of my math.  Also, note that I rounded the intermediate answer 25 N.
590280
its the 21.15,, the answer is b
321421
0.215 kL = 215L\n(1kL = 1000L)