Recent content by songoku

Sorry for late reply Thank you very much for the help and explanation berkeman, kuruman, Steve4Physics, Gavran

Sorry I got the question in form of printed paper so I don't have the link. I think the answer is (d) but I am not sure how the distribution (shape) of magnetic field inside the steel sheet would be. Is there any possibility some part of the magnetic field is redistributed by the steel sheet...

The answer key is (a) but I don't understand why. Option (a), (b), (c) are all non - magnetic material and (d) is magnetic material. Why (a) can stop the attraction while others can't? Thanks

Thank you very much for all the help and explanation haruspex, wrobel, Herman Trivilino, 256bits, Steve4Physics, jbriggs444

avoid making the assumption and use all the information provided. For this question, consideration about breaking stress was not in my head until I read your reply

I see. I need to consider breaking stress What about the formula: Young Modulus = stress / strain? Does it work under large stress because I am thinking the breaking stress of both wire is the same so the strain on the second wire must be constant and can be achieved by using longer wire? Thanks

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and...

Thank you very much anuttarasammyak

The correct answer is (C) but I don't understand why (B) is wrong. Isn't (B) also the meaning of random process? I though (B) and (C) are both correct. Thanks

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