Cap'n Refsmmat

Ah, I see. I hadn't granted anyone permission to attach files in this forum. Should be fixed now. The first error you see has me baffled. Others have reported it -- they get an error when they try to view any category. But I don't know why. For the moment, I'd suggest browsing from the forum index and only selecting forums, not categories. I'll try to track it down.

Check your first step there. 1.25 / 0.375 = 0.3? And I don't think you want to be doing division... Make sure you plug things into the correct spots in the equation I gave in post #2.

Hmm. Here's what I'd do: Figure out how many moles you'd have to do to apply the insecticide. 1.25L at 0.375M twice a week for two weeks. You can calculate (using the formula I gave above) how many moles of insecticide that is. Then, you can figure out how much concentrated solution you'd need the same way. You know the number of moles you need, and you know the molarity, so you can find the number of liters you need.

I see. And that was all as one post, not as a merged post? I'll add that to my previous bug report.

Looks right to me.

Hmm. It may be the same bug as I reported regarding merging. We'll see if it's fixed in the next version. If you continue to encounter it, try making a reduced testcase -- remove bits from the post until you come up with the smallest possible example that causes the problem. I can report further bugs to the IPB guys, as they're very quick to fix them. (Most of my reported bugs have been fixed in just a day or two...)

Here's what you need to do. If MgSO4 has a molar mass of 120.4g, and 271g are being used, how many moles do you have? Calculate that first. Next, consider the definition of molarity: [math]\mbox{molarity} = \frac{\mbox{moles}}{\mbox{liters}}[/math] You know the molarity you want - 3.75. You can calculate (like I just asked you to) the number of moles of solute you have. Solve the equation for liters and you're done.

There's a bug when you make multiple merged posts with quotes in them. Dunno if that's what you're testing. It'll be fixed in the next version of IPB.

The definition of work is [math]W=\mathbf{F}\cdot\mathbf{s}[/math], and any case where [math]\mathbf{F}\perp\mathbf{s}[/math] will result in [math]W=0[/math]. It's part of the definition of work and the dot product. I quoted my textbook already. Perhaps you have one you could refer to?

Indeed. To get the ball going the right way, there only must be force, not work, because not every force results in work. It's part of the definition of work.

No work is involved. Only force. That's the point I've been making over and over and over again. No work is required to keep an object in uniform circular motion. There is an applied force but it does no work. You can look this up in any basic physics textbook that covers circular motion.

Yes, there is. Otherwise, wouldn't the ball at the end just go in a straight line? The string's got to pull on the ball to keep it constrained to a circle, so there's tension. The string's a fixed length, though. It's not getting shorter or something.

No, the object does not move towards the center. It is rotating in a circle and stays at a fixed radius. It neither moves closer nor moves farther away. No, and I'd like to see you provide a source (university, textbook, whatever) that supports this idea. Can you find any reputable source that demonstrates that work is performed in uniform circular motion? Why would we give up on gravitation? I don't see how this discussion is relevant to it. The lack of work doesn't change Newton's law of gravitation.

That's where you're wrong. http://hyperphysics....BASE/work2.html (Georgia State University's HyperPhysics site) My textbook, Physics for Engineers and Scientists, lists the definition of work as [imath]W=\mathbf{F}\cdot\mathbf{s}[/imath], where [imath]\mathbf{F}[/imath] and [imath]\mathbf{s}[/imath] are vectors representing force and displacement (distance), respectively. Work is the dot product of those two vectors. The dot product is 0 when the force is applied perpendicular to the motion of the object, as it is in circular motion (the ball is going around the circle, while the string pulls it directly towards the center.)

I have a ball on a string that I'm spinning around. The string has a tension and pulls the ball around in the circle. The ball is traveling at a constant angular velocity. Got that? Now, is the string doing any work on the ball?

No, you haven't explained it. From what I've gathered, you're arguing that the string pulling the object in the circle is performing work. Is that right?

And I'm still applying a force to the bow so it doesn't fire, right? I still have to hold the bowstring back. But the work on the bow is complete. Thus, a force exists without causing any work. Now, consider a different situation. I have a nail, a string, and an object. The string ties the object and nail together. The object is rotating around the nail at a constant velocity, attached by the string -- much as a tetherball is attached to the pole. The object is rotating around at a constant angular velocity. The string is under tension to keep the object going in a circle. It has to apply a force on the object to keep it rotating. But the object, in the absence of friction, continues rotating at a constant angular velocity. Is the string doing work because of its force?

That's all fine. But what about when I hold the bowstring in the same position? At that point, the work is complete, but I'm still applying a force to hold the bowstring that way, aren't I? But my force on the bow is not doing any work, and that's what matters. You're making this more complicated for the sake of semantics.

Hmm. Fixed.

I know. While I deflect the bow, I am doing work, because I am applying a force to move the bowstring over a distance. Once I have deflected the bow to where I want it, I am applying a force. However, I am not applying it over a distance. It's already there. No additional work is being done. But there's still a force! So why is work necessary for there to be a force?

Sure. But once I deflect the sides as far as they'll deflect with my ten Newtons, is there no longer a force? The sides aren't moving, so there's no longer work being done, but I am still applying a force, aren't I?

So it turns out the probability of there being no birthdays is 1 when I disable the Calendar feature. Now you can calculate the odds of there being ten birthdays today...

I don't quite see your point. The deflection is finite; after a while (perhaps a second or two), the deflection will have reached its maximum and the sides will be fully opposing my applied forces. After that point, there will be no further deflection. Imagine pressing on a cardboard box. The cardboard will bend slightly, but unless you increase the force applied, it will only bend so far. But after that point, I'm still applying a force. I'm still pressing on the sides, despite the fact that they are not moving. But there's no work.

Also, the Biblical life expectancy claims do not coincide with archaeological evidence. http://www.utexas.edu/depts/classics/documents/Life.html http://www.reshafim.org.il/ad/egypt/people/index.html http://anthro.palomar.edu/homo2/mod_homo_3.htm All of these support the idea that the average life expectancy in ancient times was perhaps 20 or 30. A larger proportion of deaths at the time were due to violent causes as well.

So suppose I have a one-kilogram box. I put it on the ground. I apply a force of ten Newtons to one side, and a force of ten Newtons to the opposite side, such that the forces oppose each other equally. No work is done, because the box does not move. Does that mean there is no force? You say work must be performed before a force can exist.