PA = mg
m = PA/g
where P is the pressure inside the cooker, A is the area of the hole, and m is the mass of the weight.
m = \frac{(1.99)(10^{5})(\pi)(0.003^{2})}{(9.8)} = 0.574 kg
Homework Statement
A pressure cooker is a sealed pot designed to cook food with the steam produced by boiling water somewhat above 100 C. Consider a pressure cooker with a weight of mass m covering the only hole (which has a diameter d = 0.003 m) on top of it. What should m be in order to...
Thanks for your replies guys. I'm still not seeing it. After the substitution I get \int\frac{1}{u^{2}}du = ln(u^{2}) + C = 2ln(u) + C. I'm stumped at this point.
Thanks for your reply. Unfortunately we haven't covered integration by parts yet; we have only made it as far as logarithmic differentiation so the techniques that I know are limited. derek
Homework Statement
\int\frac{1}{xln(x)^2}
Homework Equations
\frac{d}{dx}ln(x) = \frac{1}{x}
The Attempt at a Solution
I made a u substitution, letting u = ln(x); du = \frac{1}{x}dx. The antiderivative I then get is ln(u^{2}) + C but obviously this is not the correct answer. Any help...
It looks like you represented the 2 on the right side as a common log of 100, but the left side is log base a, so that isn't very useful in this case. Since log base a ((2+x)/x) = 2, try thinking more in terms of, "a raised to the second power is equal to ((2+x)/x)."
Since you need to come up with two more zeroes, and the conjugate of (x - 5i) needs to be one of them (assuming your polynomial is to have real coefficients), use the conjugate of (x - sqrt(7)) as the other zero and it should be a little cleaner when you multiply.
eminent,
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