Recent content by wisky40

I think you shloud integrate over the whole space, so d^3x =r^2\sin\theta d\phi d\theta dr because of the spherical geometry. However, in your case it should be r^2 dr instead of dr due to the radial part. I get C=\sqrt{2 \alpha} and <r> = \frac{1}{2 \alpha}, but you should calcuate these...

maybe if you want some current it is better to use a variable magnetic field in a non-parallel position in respect to the end of the loop wire instead of electric field.

thank you. After your advice everything was really easy. I didn't look elegant, however, it was effective.

Homework Statement show that \displaystyle \lim_{n \to \infty} \left[ \left(\begin{matrix} n \\ 0 \end{matrix} \right) \left(\begin{matrix} n \\ 1\end {matrix} \right) ...\left(\begin{matrix} n \\ n \end{matrix} \right ) \right]^\frac{1}{n^2} = e^\frac{1}{2} Homework Equations Stirling's...

I think quasar987 was trying to prove the convergence on a sequence not on series. instead of writing (1+5)^n, it's better (5+1)^n or 6^n>= >= 5^n+n5^(n-1)+... or conveniently 6^n>n5^(n-1) because the term n5^(n-1) in closer to 6^n than 1+5n. finally=>...

profit=price of producing q goods-cost of producing q goods => profit= 19(q)-(.4q^2 +10q) =9(q)-.4q^2 from here you need either to find the vertex of this parabola or to derivate for its maximum.

if you want you can work it like this: x(du/dx)-xu^2=u => (u^-2)(du/dx)-(u^-1)/x=1 then use this substitution, I think this is "Bernoulli equation" z=-(u^-1) and dz/dx=(u^-2)(du/dx) => dz/dx +z/x=1. I hope this help you.

Only because you try to do something, I'm going to help. Use these forces ==> T"sin(alpha)-T'sin(alpha)-mg=0 and T"cos(alpha)+T'cos(alpha)=mv^2/r where T" is the upper tension and T' is the lower tension. Also, if you want to find cos(alpha) and sin(alpha) use your sketch

it seems you are desparate, just show me your "force equation and I help you out right know

I think tou can answer this question yourself, just check every color that you can see from that close nebula and then use the doppler effect and see if you can get any different colors or even maybe some desappear.Don't forget to use the doppler effect applied to the theory of relativity and if...

Work is defined as integral of pdv => w=Int{pdv}=Int{nRT(dv/v)},where t is constant bacause the process is isothermic => w=(nRT)In{(v_f)/(v_i)} and v_f/v_i = p_i/p_f. There are some other problems like"A 2.03 mol sample..." and then you used n=2.18 also "... at 295 K..." T= 305k... and so on

I think you can use kinetic energy and potential energy. For block A the kinetic energy (1/2)mv^2= energy dissipated on friction(onA). For block B the kinetic energy (1/2)mv^2= energy dissipated on friction(on B) +mgh' => (1/2)mv^2-mgh'= energy dissipated on friction(onB). Finally, for...

I'm going to try to help you a little bit more |(n^2+n-1)/(3n^2+1)-(1/3)|< or = |(3(n^2+n-1)-(3n^2+1))/3(3n^2+1)|< or =|(3n-4)/3(3n^2+1)|<|(3n-3)/3(3n^2+1)|< or =|(n-1)/(3n^2+1)|< |n/3n^2|< or= |1/3n|=1/3n<error => try to make the error = (1/10^k) => 1/3n<1/10^k => n>(10^k)/3. I...

sometimes it's not necessary to use calculus. |sin(alpha)|< or = 1 => the maxium value for this function is sin(alpha)=1 and the minimum => sin(alpha)=-1from here you can use these value to obtain: for "T"max, T(t)= 50+14=64 and for "T"min, T(t)=50-14 =36. wisky40

yes sir, that's correct, but be careful the answer is 530 kg/m^3 or .53 g/cm^3, and I think that is logic because the density of wood is < the density of water, and also see the proportion of h vs H ~.5 which agrees with this result.