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i am trying to solve this [t is time and T is temperature] to find the time t,
6450*Pi*(0.00185^2)*(0.0625)*836
dt = --------------------------------------------- dT
(1.2^2)*7.5 - 108*Pi*0.00185*0.25*T
between the limits 0 and 180degC
can someone help me:D
this is a bit embarassing...because i had forgotten the basics:(
how do i solve this
-5db = 20log M(f)
I see.
Thank you so much.
I dont think you did wrong , cuz 4(3,0,4)=(12,0,16)
so if i present the surface normal as (12,0,16) it is not considered as wrong?
thanks
Hi, I just got to know about this forum & I got this little problem...
I am trying to solve this cross product; 3 vertices p1=(0,0,3), p2=(4,0,0) & p3=(0,4,3) given, I need to find the normal of the surface bounded by these 3 vertices.
I was told answer is (3,0,4) but my attempts kept leading me to (12,0,16)
here's what I did.
normal = (p1-p2) x (p3-p2) = [(0,0,3)-(4,0,0)]x[(4,0,0)-(0,4,3)] = (-4,0,3)X(4,-4,-3)
normal = (12,0,16)
where did i go wrong?
Thank you
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