Recent content by ahhgidaa

i ended up with -2j-j^2 which i^2 is -1 so i finally got the answer 2-2j and i copied it exactly how it was but ure way of putting it help me with the math. idk why he writes it like that

im studying for my circuits midterm and the proff has handouts with questions and answers but not detailed answers. i can't figure out how he went from a fraction to an answer. (-j2)(2+j2)/-j2+2+j2 the answer on the paper is 2-j2 i do not know what I am allowed to do with the 2 next to...

Homework Statement 2y dy/dx = xy^2 + x y(0)= 2 The Attempt at a Solution the first step i did was take out the x on the right for x(y^2+1) 2y dy/dx = x(y^2+1) then shift things around 2y/(y^2+1) dy = x dx then integrate u substitution gives ln y^2+1 = 1/2 x^2 +...

force of tension a distance y from the bottom = m [(L-y)/L]*g tension = mgh mass per unit length times gravity

a uniform rope of length L and mass m is hung vertically. what is the tension a distance y from the bottom? my final anwer that i got was F_t(y)= [m(L-y)*g] is this correct?

1, -1

for velocity graph the max amplitude is Vmax at 3/4T and for the displacement x at 0 was 3/4T. how do you get that acc equation? i thinks that's what you were trying to have me do. but what do i think or do to go from V = -wAsin(wt) to Vmax = wA. dx/dt at x=0? then diff that. which i dnt...

well i found the equations on the next page of my book. ha. and have been looking at them and i do not see how they mathematically went from -wAsin(wt) to just wA for Vmax. I am looking at the graphs to but my math isn't strong enough to see how they could drop the trig func. but the A i got...

v=dx/dt= -wAsin(wt) and x= Acos(wt) what is the little t, time. is the time the period? and is 2 cycles per second 2 rad/sec. ? and just fmi, with the differential equation with x(t) when would i use that. when i have as specific time. I am just unsure about a lot of stuff. sry.

Motion of particle shows a max acc. of 30 m/s2 and a frequency of 120 cycles per minuet. assuming SHM determine, the amplitude and max velocity. i found the sqrt of k/m to be 12.57 even though neither was given and the frequency to 2 cycles/sec. i do not know what relations to use to work...