Recent content by blkqi

Just think of an ideal gas for a minute. Since the ideal gas equation of state, and thus pressure, is ultimately derived from the Maxwell distribution of velocities/momenta, pressure must depend upon the observers reference frame. The Maxwell distribution holds only for observers at rest with a...

The gravitational force between two massive bodies (i.e. moon and ocean or moon and human) is directly proportional to the product of the two body's masses. If the hydrosphere is, say, 10^20 times more massive than a human, then you can expect the gravitational force of the moon on a human body...

Assuming a constant rate of expansion, the approximate age of the universe is 1/H by backward extrapolation. The quantity c/H is the Hubble radius, the radius of the Hubble sphere which contains all of our causally connected, observable universe. Note that this is not a boundary of the universe...

Looks good. x=2 is clearly not a solution.

Not open does not imply closed. Most subsets of the real numbers are neither open nor closed (e.g. N, or [0,1), etc.). On the other hand, the set of all real numbers is both open and closed. If a set A is not open then there exists a point x in A such that x is not an interior point of A. For...

Hmmm.. Wolfram Alpha shows me a little different answer.. Did you forget the 1/π?This was my calculation: Integral[2*(π*sqrt(r^2-x^2))^(-1)*x^2,{x,0,r}] http://www.wolframalpha.com/

Don't square the distribution! Equation 6 is specific to 1-d quantum mechanics where dP = |ψ|^2 dx. Yet you're NOT dealing with a wavefunction. You have dP = ρ(x) dx = [π*sqrt(r^2-x^2)]^-1 dx. The expectation value is <x^2> = ∫ x^2 dP, (all space) Now take the integral. The integrand is even...

Counter example is fine if you are going for proof by contradiction. Remember the statement you want to demonstrate is that "F is not a linear transformation." In proof by contradiction you can assume that F is a linear transformation, and use a counterexample to draw the contradiction (just...

Nope. Orthogonality just requires the inner product to be zero. A good visualization (though not in Hilbert space): In Euclidean 3-space (where inner product is the dot product) orthogonal vectors are perpendicular. Normalization scales vectors to the unit length. Heres the point: two vectors...

That's the idea. You could have been a little more general in the proof, for instance: Let a \in V and b \notin V... blah blah. The fact that u is one-dimensional is a necessary but not sufficient condition for u in V. There are many one-dimension subsets of R!

Almost. You forgot to bring out the omega. It might help to ignore the scalars (constants R and omega on the OUTSIDE of your functions) when taking the derivative. Try just \frac{d}{dt}( -{\sin}\left({\omega}t\right)\hat{i}+{\cos}\left({\omega}t\right)\hat{j}) then put the factors back...

You found the derivative of r(t): v(t) = \frac{dr}{dt} = -{\omega}R{\sin}\left({\omega}t\right)\hat{i}+{\omega}R{\cos}\left({\omega}t\right)\hat{j} So find a(t)=dv/dt in just the same way.

Your method sounds good. What did you get for the second derivative of r(t)?

Any force which acts on the face of a fluid can be decomposed into perpendicular and tangential components. Just line up your horizontal axis with the face of the fluid. If there exists a tangential component (on the horizontal axis), the shear stress will cause the fluid to distort, until just...

Wikipedia cites their data for masses and populations in the table in section 2 of the Spectral Classification article. If you need a more reliable source, any good astronomy textbook should have the information you need in an appendix.