Recent content by haruspex

My approach was: When the cart has moved s from the cannon, how much water has the cannon ejected? How much of that has reached the cart? What momentum did it supply? In terms of ds/dt, how much momentum does the cart and contents have now? From that I got ##\dot...

Then you drew the triangle representing the valve the wrong way up.

I do not understand the function of the check valve here. It is drawn as though allowing flow upwards, which means admitting air if the upper piston is raised, it seems. If so, that air would rise to the top of that chamber. The net movement of water would therefore be from just above the...

In your fifth equation, the one ending …16-1)=0, check where that -1 comes from. It should be something else. Also, having obtained a quadratic in t that involves the unknown ##v_0##, bear in mind that what you want to find is t, not ##v_0##. So which should you be trying to eliminate? Btw...

I repeat: And your working, of course.

No. You have confused yourself by drawing the forces on the wedge and the block in the same diagram. Draw two separate FBDs.

The simplest way to think of it is to include the portion of the rope in contact with the pulley as being part of the wedge+pulley system. That way, you can take the forces, both magnitude F but at different angles, from the straight segments of the rope as being external forces on that system.

Really?

What has the spring to do with it? When the system is released from rest, is the spring relaxed, at equilibrium with its suspended mass, or in some other state? The homework statement you provided doesn't ask for the rate. Is this an extra question that occurred to you? I assume you mean...

Please post your result and the official result if known.

Then I misinterpreted your last diagram. I now presume it is a top view. That being so, a given movement of the piston raises twice the water than originally, so the force must be doubled.

The lower portion is not raising any water, only moving it sideways.

Just a follow-up on the equation I gave in post #18… I felt it was surprisingly inelegant, then I realised it could be rewritten as: ##f(XY)=f(X)f(Y)## where ##f(X)=\frac{E(X^2)}{E(X)^2}##.

Part of the problem is that there are two common systems which many (most) conflate. Either imply the uncertainty from the number of significant figures or state the uncertainty. "5.7” implies ##5.7\pm 0.05##. Here we are given ##5.7\pm 0.1##, so we should take both numbers as exact. Having...

It depends what you mean by the uncertainty. In many contexts you want the statistical uncertainty, i.e. standard deviation. If X and Y are independent, ##(\frac{\sigma(XY)}{E (XY)})^2=(\frac{\sigma X\sigma Y}{E(X)E(Y)})^2+(\frac{\sigma X}{E(X)})^2+(\frac{\sigma Y}{E(Y)})^2##. For small...