Recent content by Nadi Yazbeck

that's great! i have been working on this for almost a week and couldn't figure it out hahahaha thanks a lot! nadi

oh okay, so if i replace the 4 with a 2, the final result is i0(ln(h^2+R^2)-2ln(h))/4, does that make more sense? thanks

thanks, i will type from now on,it's from the inverse square law(S/(4*pi*d^4)=ir)

Is a picture fine? https://s.amsu.ng/5zxMdp9eKk6N If not let me know and i will type it out Thanks Nadi

o okay, let me know what you think about my work please, my final result is: (-R^2 * i0)/(4h^2 (h^2 + R^2)) with: R: radius of detector, h: distance between the detector and the source, i0: intensity at the source. thanks

hahahaha sure, i will try to explain the best i can: first i used Pythagoras to find the distance (sqrt(r^2+h^2)) then i used the inverse square law to find the intensity when the radius is r (ir=i0/(4.pi.d^4)) then i integrated ir*Ar (area) from 0 to R (max radius) Ar changes depending on the...

That's my work, do you think it's right? https://s.amsu.ng/6HHsQQx466QN

i meant area, sorry I'm used to french hahahahaha

i need to account for for the change in distance, but even if i find a way to calculate it, i can't figure out how to find the air to apply the intensity over(i thought about doing an integral with the circumference, what do you think?)

i was thinking about a soccer ball hahahahaah are you talking about an american football?

i'm not really sure what you mean... i will try to think about it thanks

but the detector is flat, so the power would be different for each radius... is there a way to compute a general formula? thanks nadi

can i get a hint? hahahahaha

what do you mean? does your solution account for attenuation, and the angle at which it hits the detector(the bigger the angle the fewer particles hits the detector)? thanks nadi

hahahhahha i meant bio engineering