Recent content by songoku

Thank you very much .Scott, haruspex, Gavran

For this case, is this also called induction?

If that is the case, there is discharging between the clouds so can I say there is induction between clouds? Thanks

My answer is (B) but the correct answer is (D). I don't understand why there is induction between clouds and ground. I thought there will only be separation of charges in ground (polarization) due to large negative charges in the clouds causing the ground below the clouds will be positively...

I understand. Thank you very much Gavran

(i) $$E(X)=\bar X$$ $$(-1)\left(\frac{\theta}{2}\right)+(1)\left(\frac{\theta}{2}\right)=\bar X$$ $$\bar X=0$$ Then: $$\text{Var} (X)=\bar {X^2}-(\bar X)^2$$ $$(1)\left(\frac{\theta}{2}\right)+(1)\left(\frac{\theta}{2}\right)=\bar {X^2} - 0$$ $$\theta = \frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}$$...

Thank you very much for all the explanations Lnewqban, Gavran, haruspex, Steve4Physics

How can we get this information from the question?

You mean both bat and moth will detect frequency of 230.3 kHz?

I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat...

Thank you very much for all the help and explanation kuruman, jbriggs444, Lnewqban, Herman Trivilino, Gavran, haruspex

Thank you very much for all the help and explanation jbriggs444, Chestermiller, Lnewqban, Gavran

When te straw is lifted, water level inside the straw will decrease causing the volume occupied by the air increases ans reducing the gas pressure. Is that correct? Thanks

I am not sure what you mean by "each side of the tap" but I will take position 1 to be at the 12 m above the tap and position 2 to be at the tap. Set: ##v_1=0## and ##h_2=0## $$P_1 + \frac{1}{2} \rho v_{1}^{2}+\rho g h_1=P_2 + \frac{1}{2} \rho v_{2}^{2}+\rho g h_2$$ $$P_1+\rho g...

In the first link there are several formulas for the pressure but I don't think using any of those formulas can result in P to be zero so can you elaborate more why the statement (i) is approximately correct? When deriving the equation, the pressure at max height is taken to be zero. But...