m(e)=(2/3)log(e/10^4.4)     energy release= 2*10^17   HELP?
475409
You need to define your question a bit better so I can help you. What does m(e) stand for? and i assume e denotes the exponential function?
244208
let\nl= 4h - 20,\nw= 2h + 10, and\nv= l*w*h,    then\n\nv= (4h - 20)*(2h + 10)*h\nv= (8h2-200)*h\n\nv= 8h3 - 200h
697895
E = mc^2 = 1kg * (3*10^8)^2 = 9^10^16 Joules
701562
Well this is just like sound measurement. First of all you can calculate the amount of energy released from the eq. That would be equal to 2000000*(pi*48000^2)J/s = 1.45 x 10^16 J/s\ntherefore at a distance of 1km, it would be 1.45 x 10^16 J / (pi*1000^2)sm^2 = 4.61 x 10^9 J/sm^2. \n\nRate of energy passing through 5km^2 would be (4.61 x 10^9)x(5x1000)^2 = 1.152 x 10^17 Watts = 115200 Tera Watts
134270
m2-m1 = -2.50 log(B2/B1)\nthis is the formula - \n\n?????
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
836917
WL = h/(mv)\n\nh is the planck constant\nm is the mass of the elecron\nLook for them in you book.  Plug the numbers.
515668
F = (m dot * V)e - (m dot * V)0 + (pe - p0) * Ae
539943
x+y........e=mc2..........2+2
462280
E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
889920
Yes, if it's converted to energy.  E=mc^2
372338
Note: 1.^4.4  = 1 ( 1 to the 4.4 power is what you have written.\n\ne/(1.^4.4) = e/1 = e\n\nlog(e/(1.^4.4)) = log(e/1) = log(e) = 1\n\n(2/3)log(e/1^4.4) = 2/3\n\nIs your question stated correctly?
762662
Minimize the energy equation by finding its derivative, setting that equal to zero, and solving for F. This will find the local minimum(s) and/or maximum(s).\n\nThe derivative of E is 0.25 - 3.4/F^3.\n\nSetting this equal to 0 and solving for F gives us:\n\nF = 2.387. Plug 2 points (1 less than F and one greater than F) into the derivative equation to find out whether F is a maximum or minimum. Since E is negative when F < 2.387 and E is positive when F > 2.387 the point is a minimum.\n\nThe bird can minimize its energy expenditure if it forages for 2.387 hours per day.\n\nI hope this helps, good luck!
1033843
e=mc^2\n\ne (energy)=m (mass) times c (the speed of light) squared
188896
K = mv^2 = (10kg)(4.5 m/s)^2 = 202.5 J\nI would suggest you try doing your homework, asking for all the answers will only hurt you in the future.
246299
h= (C*I/A+T/k)*z*log(.7*y)
739092
PE at release point = KE + PE at the top of the loop.\n\nmgh = 1/2 m v^2 + mg 2R   the 2R is he height of the loop.\n\nSolve for KE\n\nmgh-mg2R = KE = mg(h-2R)\n\nNow for the second one I believe you have to use Net F = ma\n\nAssume the normal force is 0, because the car is barly staying on the track. so Fgrav = m v^2/R\n\nso mg = mv^2/R\nso g = v^2/R\n\nNow going back to the energy equations, mgh = 1/2 mv^2 + mg2R\n\nm's cancel so gh= 1/2 v^2 + g2R\n\nsolve for h and you have h =    (.5v^2 + g2R)/ g  make that into 2 fractions and you have \n.5v^2/g  +  g2R/g  and v^2/g from the equation above = R\n\nso .5R + 2R = h  so 2.5R = h\n\nHope thats right!