Recent content by sapta

You can also divide 2n + 6m \equiv 2( mod\ 10) by 2 on both sides to get n + 3m \equiv 1( mod\ 5) n + m \equiv -2 (mod\ 10) \implies n + m \equiv 8 (mod\ 10) \implies n + m \equiv 3 (mod\ 5) Now solving the mod 5 congruences, gives m \equiv 4 (mod\ 5) which means 2...

I am 1st year B.E. ,stream CSE.

Please suggest a good book on Mechanics covering the following topics- Newtonian Mechanics – difficulties to handle coupled equations, Constraints (both time dependent and time independent), Degrees of freedom, Generalised co-ordinates, Generalized force, potential and kinetic energy...

A book of mine has the following questions with no answers.So I am not sure if I am right or if the proceure is correct and the best.So,please help- 1.For a given real number a>0,define A_n=(1^a + 2^a + 3^a +...+n^a)^n and B_n=n^n(n!)^a for all n=1,2,... Then a. A_n<B_nfor all n>1...

I understand(to some extent).The confusion was probably arising because of the definition of luminous intensity given in my books which say- The amount of light passing through a unit solid angle per second is the intensity of the source emitting the light. I think the part in italics...

consider the forces acting along the incline.when you consider mg acting downward,u cannot additionally consider g.So,do the following- Resolve mg along the incline which would be mgsin30.So,net force acting along the incline=mgsin30-F where F is frictional force opposing the movement...

yes,this should be the answer in farad. If you are not allowed to use the formula directly,then it would be better to consider two capacitors in series rather than to deduce the formula.

Okay,1 lm is emitted each second for each solid angle,but isn't that intensity (which is light emitted per second through a unit solid angle)?.so, how can lumen be a unit of flux?candela is measured for 4pi sd solid angle,so candela should be the unit of flux:confused:

1-cos theta =(sin theta/2)^2

area is (.27 m)^2;also d=1.85 x 10^-4 m.

consider two capacitors in series each having distance d = half of 0.37 mm.The first has dielectric k=4.93,second has k=10. Or you may use the following equation- C=EA/[d/k + d'/k'] here d=d'=half of 0.37 mm

hmm,so flux is power. never thought it that way.Okay,from now on I shall define lumen in terms of watt.:cool: But I am still confused over the definitions I gave.Take a look at the following relations that I have deduced from the first two posts in this thread- 1)Luminous flux=4pi ><...

1. What is luminous flux and what is luminous intensity? How are they related? 2. What is the definition of lumen (unit of luminous flux)? How is it related with candela? There are a lot of definitions in my books but I am a bit confused, particularly over the definition of lumen. One...

I am still waiting for an answer

O.K. as you said,I deduced the potential energy difference which is GM'm/r. So,mv^2/2=GM'm/r -->v=(2GM'/r)^1/2 Therefore,v3=v-(GM'/r)^1/2. But this is nowhere near the answer.I have not considered any influence of the Earth which is most probably wrong.So,where & how should I take the Earth...