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i can help u in datThanks sameen I just didn't know this formula!
And hey can you help me with number 14 as well?
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i can help u in datThanks sameen I just didn't know this formula!
And hey can you help me with number 14 as well?
where is da question ?
somwhere abovewhere is da question ?
M/J 2005 Q7cwhere is da question ?
how did u reach B0/Bn = mo/mn ??
edited my reply checkCan u inbox me ?
Bo= mov/qr nd Bn= mv/qr ..v , q nd r are same so they cancel each other..how did u reach B0/Bn = mo/mn ??
Calculate the path difference between the two waves, which is .15m. Now use the wavelength from part b and divide the path difference with it.papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s13_qp_21.pdf
Q5 (c) how do we know that a maxima is observed at point P?
they said to use data from Fig.8.1 given in the qts not the values from data sheethttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_43.pdf Q8b(i) why did they take the rest mass different for proton and neutron when in the data sheet they have given only one value
Calculate the path difference between the two waves, which is .15m. Now use the wavelength from part b and divide the path difference with it.
.15/0.025 = 6 waves. This means that wave S2 has travelled 6 complete oscillations more than S1 before reaching the screen. Furthermore, as the number is a whole number, we know that there will be constructive interference. Thus there is a maxima at P.
You can also calculate this by dividing the paths of both the waves with the wavelength. One would be 30, the other 36
U2thank you stay blessed
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