I just noticed this as well. BTW I was just wondering the way I got rid of my absolute values was correct right, regardless of the general problem i stated?
if y(x) > 5 then |y-5| = (y-5)
and if y(x) <= 5 we get |y-5| = -(y-5) ?
then we get two solutions that i posted above again? Am i missing something here
Homework Statement
\frac{dy}{dx}\:+\:ycosx\:=\:5cosx
I get two solutions for y however only one of them is correct according to my online homework
(see attempt at solution)
Homework Equations
y(0) = 7 is initial condition
The Attempt at a Solution
\int \:\frac{1}{5-y}dy\:=\:\int...
I understand that the velocity should decrease, but isn't acceleration in the same direction as the drag force so they should have the same signs in the force equation? This is quite non - intuitive for me.
Homework Statement
A 500 kg boat moves with an initial speed of 20 m/s turns off it engine and slows down due to the drag force. If this force has magnitude F= 15.3v^2 find the boat speed after 12 seconds.Homework Equations
∑F = ma
dv = adt
The Attempt at a Solution
taking the side opposite...
Well it only encloses the charge Q not q. So is the electric field calculated at P1 only due to Q? and the electric field due to q is zero at point P1 because of gauss's law?
is this correct?
thanks.
Sorry, Let me rephrase my question because there may be some misinterpretation. is the electric field that the book found using gauss's law the total electric field or just the electric field due to Q (the point charge only).
I believe it states that the total electric flux out of / into a closed surface is equal to the charge enclosed divided by the permittivity. it pretty much relates the electric field at points on a closed surface to the
net charge enclosed by that surface.