thnx again. i understood ths
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They have already provided you with the electric field strength and the charge and are asking you what the additional charge should be to have the electric field strength as 2x10^6can some one explain this please?
Use the fact that <c^2> is directly proportional to the kelvin temperaturecaView attachment 62084ca some one explain b part ii
thank youu sooo sooo much bro. thanks a lot . i just got itUse the fact that <c^2> is directly proportional to the kelvin temperature
Cross multiply: 1.9 x 10^6 ------------> (32+273)
? -------------> (80+273)
Once you get the mean square speed at 80 (that is, '?'), square root it to get root mean square speed
Hey guys, how many significant figures shoud our answers in P4 be?
Same as the least number of s.f in the raw data?
For the first one, if you see the equation for electric potential which is V = Q/(4*pi*e*r) you can easily see that V is inversely proportion to 'r'. So when 'r' is double, V is halved. The electric potential remains same (constant) till R. Then after that when R is doubled, V is halved, so the curve must pass through (R, Vs) & (2R, 0.5Vs).Someone pleeassse explain these sketches. MS says smooth curves and I'm not getting how!
View attachment 62119
Yeah thanks..For the first one, if you see the equation for electric potential which is V = Q/(4*pi*e*r) you can easily see that V is inversely proportion to 'r'. So when 'r' is double, V is halved. The electric potential remains same (constant) till R. Then after that when R is doubled, V is halved, so the curve must pass through (R, Vs) & (2R, 0.5Vs).
Do the same thing with the next graph.
A truck of mass 500kg moves from rest at the top of a section of track 400 m long and 30 m high, The frictional force acting on the truck is 250 N throughout its journey.What is the final speed of the truck? Help please. this is from october november 2016 varient 1 answer should be 14ms-1
I am not quite sure, but our teacher told us that for such qs measure time and number of oscillations performed during that time and then divide the total time taken by the number of oscillations. And repeat the experiment atleat twice and then take an average of the times taken.won't it be too fast to measure period of such oscillations?
how do we do it then?View attachment 62163
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