Recent content by sanhuy

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?

both solutions work, but my webwork says the first one is wrong

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

Since 2e^−sinx>0 you can drop the absolute sign correct? then it becomes y=5−2e^−sinx?

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...

THANK YOU! I Get it now!

Oh I see. So, just to confirm, v hat is the unit vector that points in the direction of the velocity correct?

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...

Thanks guys.

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.

wait, is the electric field inside of a shell zero due to symmetry if the charge is uniformly distributed on the spherical shell?

No, not really. :\

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.