Given the triangle above where V < v'_{1}, prove that the \[ v_{1}=V \cos(\psi)+v'_{1} \cos(\theta - \psi) \]
It is said that v_{1} is equal to the sum of the orthogonal projections on v_{1} of V and of v'_{1} and that is precisely the expression that show. But I couldn't see how to make the...
I think yes, if I rewrite like
\[ \mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2} \]
but I have \[ (s+\lambda)^2-\omega^2 \] and not \[ (s+\lambda)^2+\omega^2 \]
The table of Laplace transforms lists that \[...
Does anyone know how I can relate the diffraction in a circular crack to the thermal expansion of that crack?
Something that I relate the gap radius with the distances between the light and dark fractions of the diffraction figure.
A beam of length L with fixed ends, has a concentrated force P applied in the center exactly in L / 2.
In the differential equation:
\[ \frac{d^4y(x)}{dx^4}=\frac{1}{\text{EI}}q(x) \]
In which
\[ q(x)= P \delta(x-\frac{L}{2}) \]
P represents an infinitely concentrated charge distribution...