Recent content by Dopefish1337

Yes, ln(e)=1. I kindve messed up my hint there and ended up inadvertantly giving answer instead. (e^0 is 1, but that's not relevant here.) Remember ln is just log base e, which 'undoes' powers, so you you're looking for the power of e that equals e in this case. (e^1=e^ln(x) => e^(lnx) is...

e to the what equals 1?

Really? I thought 1/x was the derivative of lnx, not xlnx.

Question closes in around 5 hours. Anyone?

It's been awhile since I've done stats, but doesn't the t value depend on your degrees of freedom, which in turn depends on n?

Erm, they are the same on both sides. You have C, H, N, and O on the left, and C, H, N, and O on the right. Remember you can don't need to multiply entire sides by a number, you can just multiply each product individually, so you can have 3N2 on the right but 6 CO2 for example.

Homework Statement The concentration of retinol (vitamin A) in serum was determined by HPLC with UV absorbance detection. A 1.00 mL sample of serum was placed in a 10 mL test tube. To precipitate the proteins, 800.0 microL of ethanol was added, and the test tube was agitated. Next, a 2.00 mL...

I suspect if your question is similar to mine, then you'd want an angle in your formula for normal force. If you plot the function of B as a function of theta, you can just manually look through the table and find where B is minimum.

What expression do you have for frictional force? (or are you saying that the frictional force *is* ILBsin(theta), as opposed to simply be equivalent?) Once you have an expression with only B and theta as unknowns, you can rearrange to get B as a function of theta. Once you have that, there's...

Got it now, think my cos and sin were reversed in my previous post, but I fiddled with things until it worked, and it has. Thanks for the help. :)

What would the resultant equations be that lead to that conclusion? Fn=mg-I*l*b*cos(theta)? So that I*l*b*sin(theta)=(mg-I*l*b*cos(theta))*.503? Or did I mess up my expression for Fn?

I'm toying with the idea that perhaps it might be possible that if the field were on a slight angle, the rod itself might feel a slight vertical force, thus causing the normal force to decrease and in turn the frictional force. However, I don't think that would actually decrease the field...

If you used velocities instead of momentums for your base and height, then yes. If you used momentums, there's no need to multiply by mass.

If the vectors you use are momentum vectors, than the hypotenuse will be momentum as well. (It would be possible to do it with velocities as well, but that's more work.)

One of your vectors will point up like "|", and the other sideways like "_", so you have the base and height of a right angled triangle. You want the hypotenuse.