Recent content by Gavran

All right. Based on post #19, there are the next steps. $$ \frac{d}{d\phi}(8+\frac{8}{\tan\phi}+2\tan\phi)=0\implies(-\frac{8}{\tan^2\phi}+2)\cdot\sec^2\phi=0\implies $$ $$ -\frac{8}{\tan^2\phi}+2=0\implies\tan^2\phi=4\implies\tan\phi=\pm2\implies\phi=\pm63,435^\circ $$ ##...

I suppose that you use the Pythagorean theorem here. (see below) $$ \sqrt{(a+2)^2+(b+4)^2}=\sqrt{b^2+2^2}+\sqrt{a^2+4^2}\implies0=0 $$ So, there is not a unique solution. The minimum of ## S ## where $$ S=8+\frac{8}{\tan\phi}+2\tan\phi $$ you can find by equating the first derivative ##...

And what is your answer? Einstein or Newton? Please choose one, not both of them.

When we speak about electrostatic induction, we speak in the context of charges of regions of the object, not in the context of charges of the overall object. The object can be neutral, but its charges cannot be evenly distributed. The polarization in this case is caused by what you have...

Newton His work was the groundwork for later discoveries.

Top three: Chernobyl, the Bhopal Disaster, and the Titanic.

Only rotations are considered to be the same.

(i) Your work is correct, but your notation should be clearer. should be $$ \hat{\theta}=\frac1n\sum_{i=1}^{n}X_i^2 $$. (ii) I do not understand how you get this . By following steps $$ E(\hat{\theta})=E(\frac1n\sum_{i=1}^{n}X_i^2)=\frac1nE(\sum_{i=1}^{n}X_i^2)=... $$ you should get this $$...

By replacing humans with artificial intelligence, this forum would lose its soul, and some kind of emptiness would be left.

Your solution is okay. You can check it at https://en.wikipedia.org/wiki/Doppler_effect#Radar.

Okay. ## c^2=a^2+(2R)^2-2\cdot a\cdot2R\cdot\cos(\angle ADC) ## ## c^2=x^2+b^2-2\cdot x\cdot b\cdot\cos(\angle ABC) ## I want to offer one more solution to show that the decision between ## c ## or ## d ## is irrelevant. ## \cos(\angle BCD)=b/(2R) ## from The definition of cosine in the...

You are right. There is one more approach which is based on the figure from the post #39 and it is simpler than the one I have provided in the post #43. ## \cos(\angle ADC)=a/(2R) ## from The definition of cosine in the context of a right-angled triangle ## \angle ADC+\angle ABC=180^\circ ##...

Yes it does and this holds in theory. There is not an idealized situation in theory when PB=0 and the syphon works. hB=Patm/(ρg)-vB2/(2g) is the height of the syphon which can not be reached because at that height PB=0 and the syphon will “break”. This value limits the height of the syphon. The...

The formula is correct.

One more approach based on https://en.wikipedia.org/wiki/Cyclic_quadrilateral and the cosine theorem. ## \angle FGB=120^\circ\implies\angle FEB=60^\circ ## (Supplementary angles) ## \cos(\angle FEB)=(c^2-7^2-7^2+11^2)/(2\cdot(11\cdot c+7\cdot7))\implies c=13 ## (Angle formulas) ## \cos(\angle...