m(e)=(2/3)log(e/(1.^4.4)    energy released =2*10^17?
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?
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
968058
Type 1.16 log / 4 / 1.0125 log =\nThis is 3 approximately (2.9869...)\n\nTh.
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?
711054
It's two.  Energy goes as the square of amplitude.
445873
K.E is given by m*(v^2)/2\nMomentum p = mv\nSquare both sides you get p^2 = m^2*v^2\nNow divide both sides by 2m and you get p^2/2m = m*(v^2)/2\nClearly the right hand side of the equation represents K.E\nso the answer is K.E = p^2/2m
815275
Using the quotient rule,\n\nlog(2x-1) - log(x-3)=1\nlog((2x-1)/(x-3)) = 1\n\nThen change to exponential form:\n\n10^1 = (2x-1)/(x-3)\n\nthen solve:\n\n10(x-3)=2x-1\n10x-30=2x-1\n8x=29\nx=29/8
840583
F = ma\n\nF = m(Δv/Δt)\n\nF = 38000(17-0/20)\n\nF = 38000(17/20)\n\nF = 3.2 x 10^4 N
827623
F*m = μ*Joules→ m = .25*40000000/1000 = 10000m = 10 km
819160
em energy radiated is proportional to T^4, so it would be : ((273.15+9627)/ (273.15+5227))^4 =10.5
745284
well, work and energy is about the same\n\nso w = fd = mgh\n\nm = 1000\ng= 9.81 m/s^2\nd = 146\n\nw = 1000*9.81*146 (N/m)\n\nplug it in there you go
737575
4=2.3log(R+3000)-5.1\n9.1=2.3log(R+3000)\n3.9565=log(R+3000)\n10^3.9565 = R+3000\n9,047=R+3000\nR= 6047
754118
E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
649603
To say its an equation involving the relation between energy possesed by a mass
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
772997
A/B=e^(-t/5)\nln(A/B)=-t/5\n-t=5ln(A/B)\nt=-5ln(A/B)
153995
F=m*a\na=F/m = 3.5 x 10^7N/(2.0x10^6kg) = 17.5 m/s^2 or 1.8g