Did those scientists find any evidence of any sexual behaviour between them?
As it is said, since they are in captivity, they could just be shring their grief. Homosexuality is not a natural behavior between animals at all.
hhegab
Can you put like , first case is true if f(x) is even and if such and such....
I shall study your answer. But from my first reading I need more.
I need a condition to apply to f(x) so I can use each of the above properties.
hhegab
Peace!
I want to know the conditions that must be satisfied by a function
f(x) for any of the following two cases to be true (each case independent from the other);
1- \int^a_{-a} f(x) dx = 2 \int^a_0 f(x) dx
2- \int^a_0 f(x) dx = \int^0_a f(x) dx
They gave me confusion when...
Peace!
I wanted to find the elecric field of the rod to the left and then find the force element acting on the right rod.
I got;
E=\frac{k_eQ}{d(2a+d)}
now,
dF=E\ldot dx
then what shoudl I use for integration limit?
hhegab
Thank you all,
the number 42; I am not sure I owe anything but my good company :)
I really missed this place. I feel I am living again when I came back...life is boring, but some can make it a nice experience....enjoy it here!
hhegab
Peace!
I have a problem when attaching pics to my messages. I can't know about the pixel size of my attached file. How can I know it? And how can I reduce it if it is larger than the limit?
hhegab
PS.
I am not sure if this is the correct place of my post, so please forgive me if it is...
Peace!
The problem is that I can't grasp the answer!
I have tried with the integration but I use different limits for integration. I shall post the answer if you want to.
hhegab
Peace!
I am supposed to find the electric force acting on the right rod due to the rod on the left. each is of length 2a and b > 2a. Each rod carries a charge +Q. I have difficulty in finding the integration, especially its limits.
hhegab
Peace!
I did the problem twice and I had the same answer as that of A M.
Either you gave us a wrong input somewhere, or that the answer you have is not accurate...Please check the problem and the inputs in it.
I have aslo tried to solve it assuming that a charge Q was divided equally...
Peace!
The field lines are perpendicular to the surface, and directed either inward or outward according to the sign of sigma, when they added (vectirially) they will give a zero field.
Also, even in the presence of a field inside the conductor, when we put it an an external elecric field...