Recent content by Vir

I thought gases and liquids had the same properties? The question specifically asks me to figure it out using these data.

Homework Statement I have 1.5 kgs of silicon with temperature 40 degrees celsius. It is dropped into 3 kgs of water holding temperature 25 degrees celsius. The system is heat isolated from the environment and the final temperature of the system is 26.2 degrees celsius. I need to find the...

So I checked my book: The potential dV(x) at P from dM, that is now a small part of the ring, at distance S is given by: \begin{equation} dV(x) = \frac{-GdM}{S} \end{equation} \begin{equation} dM = \frac{M}{2 \pi R}Rd \theta \end{equation} So \begin{equation} V(x) = \frac{-GM}{2 \pi...

I don't know. My integral gives me the right answer for R>r: \begin{equation} V(r>R) = \frac{-GM}{2rR} \int_{r-R}^{r+R} ds \end{equation} Simplifies to: \begin{equation} V(r>R) = \frac{-GM}{r} \end{equation} I think the point is that when r<R then the rings both to the left and right of P...

Right. So what do I do with this?

Then I don't know. You asked about the magnitude of the remaining force, isn't that the horizontal component of the total force?

That would depend on the mass at P right? In any case it's \begin{equation} F = -\frac{GMm}{(r-R Cos \theta)} \end{equation}

Yea, ok. I was thinking since each point on the circle has a point on the opposite side where the perpendicular component of the forces would cancel out the net force would be parallell to the axis through the center of the circle and P. Not so?

When the ring is to the right of P the force points left, and vice versa.

Ok. So \begin{equation} dV(r<R) = \frac{-GdM}{S} \end{equation} \begin{equation} dM = \frac{M}{4\pi R^2}d \theta R Sin \theta R 2 \pi \end{equation} \begin{equation} dM = \frac{M}{2}Sin \theta d \theta \end{equation} \begin{equation} S = \sqrt{R^2+r^2-2Rr Cos \theta} \end{equation}...

Yes, a hollow sphere. I'm integrating small circles of the surface of the sphere. I guess this image describes it well.

I'm trying to arrive at the gravitational potential at a distance r < radius of sphere(R). The integral is \begin{equation} V(r<R) = \int \frac{-GM}{2rR}ds, \end{equation} but I'm not sure how to select my integration limits.

Homework Statement So I'm calculating the gravitational potential of a sphere at at point P. R = radius of sphere, r = distance from center of sphere to point P. I'm looking at two scenarios; r > R (1) and r < R (2). So I have the following integral: \begin{equation} V(r) = \int...

Thank you for pointing this out, as it was the part I was missing. The new velocity vector is at a distance R from my reference point, which gives me the equation: \begin{equation} m R \cos θ V_0+ Iω_0 = mRV_1 + Iω_1 \end{equation} which simplifies to \begin{equation} ω_1 =...

So, as the ball touches the incline, the angle between the two contact points is θ which gives me \begin{equation} x = R - R cos θ. \end{equation} wR = v I'm not exactly sure what you mean. The velocity is at an angle θ of the horizontal.