Recent content by bigplanet401

Is height really the only thing that matters here? Intuitively, this just doesn't make sense to me. How can a small straw of liquid (say 10 cm high) exert the same pressure at the surface of the barrel as, say, a huge vat that is just as high?

Homework Statement A small tube is connected to the top of a larger one and the whole thing is filled with water. The small tube has height a and the larger tube has height b. What happens to the pressure at the bottom of the larger tube as (1) a is varied, and (2) a is held constant but the...

The integrand is not continuous over [a, b], so you can't use the fundamental theorem of calculus (directly). Try the substitution u^3 = x and write the integrand in terms of u. Make sure you change the limits of integration, too. Then try v = u - 2. Then things will look a little more obvious.

Hmm...yes, it does look like the second term in the denominator stays small for depths up to 1000 m. At 11000 m (the Marianas trench is about as deep), the denominator is 1 - 0.049, which to me means the model ought not to be used here. Thanks! That was not obvious to me at all.

Homework Statement Show that the density of water at a depth z in the ocean is related to the surface density rho_s by \rho(z) \approx \rho_s [1 + (\rho_s g/B)z] where B is the bulk modulus of water. Homework Equations B = -V (dP/dV) B = rho (dP/d rho) 3. The Attempt at a Solution I've...

Thanks for the hints. After thinking about it more, I came up with the following sum for the torque: N = \sum_{i=1}^N \; \left(D - \frac{Di}{N} \right) \rho W g \frac{D}{N} \frac{Di}{N} where D is the depth of the water, N is the number of slabs (each of thickness D/N), W is the width of the...

Homework Statement A body of water of depth D sits behind a vertical dam. The water and dam are in static equilibrium. Calculate the torque on the dam due to the water about an axis at ground level (that is, a depth D below the surface of the water). Homework Equations N (torque) = r x F The...

Oh...minus sign! That makes since because the force of gravity is directed to the center of the Earth, in the -r^ direction. Thanks!

So far so good...sine and cosine for the right!

Homework Statement Why is the gravitational potential energy of a ball a distance r from the center of the Earth negative? Homework Equations U_\text{grav}(r) = - GMm/r [/B] (To me, this makes sense because gravity is an attractive force and bodies will want to minimize the distance between...

If x = f(t), dx = f'(t) dt. I understand that part. But in \int y \; dx isn't y = y(x) a function of x? We'd then have y = y(x) = y[x(t)]. How can we just let y = g(t) and get the resulting expression in t?

Homework Statement Use the parametric equations of an ellipse, x = f(t)= a cos t and y = g(t) = b sin t, 0 <= t <= 2 pi, to find the area that it encloses.Homework Equations Integral for parametric equations. The Attempt at a Solution A = \int_0^{2 \pi} g(t) f^\prime(t) \; dt = \int_0^{2...

Homework Statement Why does \int_a^b \, y \; dx become \int_\alpha^\beta \, g(t) f^\prime(t) \; dt if x = f(t) and y = g(t) and alpha <= t <= beta? Homework Equations Substitution rule?The Attempt at a Solution I'm not sure how y = y(x) in the integrand turns into g(t). Isn't y a...

Maybe one way of looking at it is to ask if \sqrt{4} = -\sqrt{4} and, if \sqrt{4} = \pm 2, +(\pm 2) = - (\pm 2) which is a true statement. But this would mean -x = x with x nonzero, which is false. So this means \sqrt{4} has only one value. Is this kinda sort of right?

But (-2)^2 = 4? The square root of 4 is 2.