Recent content by cheerspens

Re=unitless D=meters n=Hz=s-1 \rho=kg/m3 \mu=m2/s Where do the kilograms cancel out in this equation? I was correct to convert cm to m and g to kg right?

How do you get the grams from the density to cancel out in the Reynolds number equation?

Homework Statement If the Reynolds number is 4000, what is the Power (P) required in units of watts at a shaft speed (n) of 0.03 Hz? Assume the system contains acetone, with a kinematic viscosity of 0.419 stokes. The density of acetone is 0.785 grams per cubed centimeter. Us Curve B in the...

Homework Statement How large a surface are in units of square feet will 1 gallon of paint cover if we apply a coat of paint that is 0.15 centimeters thick? Homework Equations There is no shape given that the paint must cover to know which volume and area equation to use. There must be...

Homework Statement The picture for my question is attached. I labeled on the picture, which is a metal object, where I thought the distribution of charges on the surface would be if it has a net negative charge and the electric field lines near the surface. At what point on this object...

Homework Statement The questions are based off of the diagram I attached. First, I had to find the strength of the electric field due to each charge at (0 cm, 8 cm). Now I need to find the total electric field (both magnitude and direction) at (0 cm, 8 cm) and the total electric potential...

(q1=+6.00 nc) <---------------- d=0.600 m ----------------> (q2=-3.00 nc) -------->(q3)

So you set the magnitude of q1 on q2 equal to the magnitude of q3 on q2 and solve for the distance (d-.6)?

I'm not sure if you took the time to actually solve it (I wouldn't blame you if you didn't) but for my final answer I got 6.88nC. If the book's answer is 7.2nC would it be safe to say this is accurate?

Ok I've made sure that I account for both q's. In order to solve this, do I have to break the vectors into components and set them equal to each other?

sin(180) = 0 cos(180) = -1 sin(270) = -1 cos(270)=0

I did and that's how I got 2.7x10-9\hat{x} + 0\hat{y} Was my set up right and I went wrong somewhere in the math? I was always thinking I set it up wrong or wasn't using the right values.

FCB=[(9x109|(6x10-9)(5x10-9)|) / (.32)][cos 180]\widehat{x} + [(9x109|(6x10-9)(5x10-9)|) / (.32)][sin 180]\widehat{y} FAB=[(9x109|(-3x10-9)(5x10-9)|) / (.12)][cos 270]\widehat{x} +[(9x109|(-3x10-9)(5x10-9)|) / (.12)][sin 270]\widehat{y}

Homework Statement Three charges are arranged as shown in the picture I attached. Find the magnitude and direction of the electrostatic force on the charge at the origin.Homework Equations I know that Coulomb's Law must be applied.The Attempt at a Solution I've drawn a force diagram with FCB...

The angle I got isn't even correct. According to the book it should be 77.5 degrees below the -x axis.