Recent content by tahmidbro

I attempted using f = 1/(2pi x sqrt l/g) For Earth I found the value of length to be 0.0276m. Then I substituted the value in the equation, putting (1/3)g instead of g, to find the value of f in Mars. My answer is C. I am confused. Please help me.

m= 0.014 kg E = 200MeV mol of Uranium x avogadro constant = no. of mol of uranium nuclei So , energy released = 0.014/235 x 6.02 x 10^(23) x 200 MeV = 7.17 x 10 ^21 MeV Hurrah! A friend of mine helped me to do it. Anyway, Thanks to you all! :-)

Actually, ''dinitrogen'' is not part of the second question. It was part of the answer from solution book.

As far as I know, Uranium 235 can absorb a neutron and can undergo fission, not U- 238. Probably this is why they have given % of uranium 235 in the question.

the answer in the solution book is 29K which only comes if I use mass for only one atom. ( They did not show any working ) My attempt: 1/2 x (1.67 x 10^(-27)) x (355)^(2) = 3/2 x 1.38 x 10^(-23) x T T = 29.48820652 K The confusion arises when I tried the following question: Q. Estimate the...

There are (2000/235) x 6.02 x 10^(23) = 5.1234 x 10^(24) atoms in 2kg of fuel. If 0.7% of these fission, 5.1234 x 10^(24) x 0.7%= 3.586 x 10^(22) atoms will undergo fission and if fission of each atom releases 200 MeV, then 3.586 x 10^(22) x 200 x 10^(6) x 1.6 x 10^(-19) = 1.1476 x 10^(12)...

Summary:: Calculate the amount of energy in joules generated from 2 kg of uranium fuel if the uranium 235 represents 0.7% of the metal and every fission releases 200 MeV. Hi! I am stuck in this question from my exercise book : Q. Calculate the amount of energy in joules generated from 2 kg...

Okay , Thanks for the help!

Sorry for forgetting to upload the 'worked example' along with the question My bad :-d

Now I understand. Thank you!

w = 2pi/8 =0.785 A = 3/100 metre . For maximum velocity at 'z', sinwt = 1 v= Aw = 0.785 x 3/100 = 0.0235 m/s But the answer from the book is 6 m/s. How do I get that?

Will you please share your calculation here? are you using v = -Aw sin(wt) ? ( this equation is not in the chapter of the exercise ) well, I am studying by myself. I do not have any instructor or a teaching assistant.

See the question : https://www.thestudentroom.co.uk/att...hmentid=978958 The mark scheme/answer : https://www.thestudentroom.co.uk/att...hmentid=978956 I have got the answer to the vertical height gained = 1.355 m. No problem. But not the value of the percentage difference. Their value : 23%...

Yes. I have drawn a tangent with the curve at 'z' and calculated the gradient. Gradient = ( 3+3 )/(7-5) = 3 cm/s = 0.03m/s. But the answer is 6m/s. Will you please tell me how?

How is the answer to 3 (d) is found?