Recent content by szynkasz

If \theta is the angle between \vec{k} and \vec{G} then the angle in the law of cosines should be 180^o-\theta

Because it is 35\,V: V_t=V_{oc}=10\cdot\frac{14j}{14j-10j}=35

Yes, it is correct.

\vec{u}\circ\vec{v}=|\vec{u}|\cdot|\vec{v}|\cos\angle(\vec{u},\vec{v})=-1\Rightarrow \cos\angle(\vec{u},\vec{v})<0\Rightarrow 90^o<\angle(\vec{u},\vec{v})<270^o the same direction: \cos\angle(\vec{u},\vec{v})=\cos 0^o=1\Rightarrow \vec{u}\circ\vec{v}\ge 0>-1 so not B...

Assume that light is emitted uniformly into a sphere. Its radius is the distance to observer.

Calculate angles of the triangle ##CDE## in terms of ##\angle BAC## Then use the sine formula to relate ##|CE|## to ##|CD|##

Improper integral ##\int_{-\infty}^\infty f(x)\,dx## is converged if both integrals ##\int_{-\infty}^af(x)\,dx## and ##\int_a^\infty f(x)\,dx## are convergent. Cauchy principal value is a limit ##\lim\limits_{a\to\infty}\int_{-a}^af(x)\,dx## and it may exist even if ##\int_{-\infty}^\infty...

Can you give an example of a circuit you have problem with ?

Consider ##\lim\limits_{t\to 0}y(t)## in dependence on ##a##

Yes, you have ##\int_0^tf(\tau,t)\,d\tau## in the integral instead of ##\int_0^tf(\tau)\,d\tau## and the formula for integral doesn't work. This is convolution so: ##L\{1*t^3\}=L\{1\}\cdot L\{t^3\}##

Unfortunately it isn't correct. For convolution we have: ##L\{f*g\}=L\{f\}\cdot L\{g\}##

If you have ##dS## it is rather surface integral, not line one.

Mhorton91, how did you get: ##-\frac{y}{yz}=x-x## from: ##-y=xyz-x## You can't divide one term on the right side by ##yz## and leave the other unchanged.

Hint: put terms with "x" one the one side of equation and terms without "x" on the other

The 80 N force is decomposed into two vectors and moments of both vectors are added, it makes calculations easier.