s pepperchin

I don't know if you still need help with this but the first part is just vector addition, the second part is solving the y direction to go a distance W and the third part is multiplying the downstream velocty from part one by the time found in part two. If you need to see the actual proof let me know and I will post it.

I bet that last post made you feel better about yourself, didn't it. Im just trying to offer another method for posting equations that makes them easy to read.

its not about using powerpoint its about using the equation editor which allows you to make the equations look the way they do when you use LaTeX but equation editor gives you dropdown boxes to choose from so its more user friendly. Its a little extra work that makes it easier for others to understand exactly what problem your trying to solve.

How did you set up the problem? I would determine the equation for the area of the tank and pipe as a function of height, then I would just integrate over the limits from zero to the top of the tower. This will help you get started if you need more help post again and I'll give you some more direction. Also it would be nice to see your work. If it is wrong than we might be able to tell you where to correct instead of starting over every time.

If you don't want to learn LaTeX the esiest way to import problems is to use PowerPoint you start with an empty slide, no textboxes or anything, and start by drawing a white box over most of the slide. This is so when you save it it will be visible. Next use the equation editor, to use it go to the menu and choose insert object and equation editor is one of the choices for object type. You can also place text boxes on the frame for writing out problems or you can draw diagrams. Once you have it the way you want use the arrow to draw a selection box outside of the white box. When you do this everything should be selected. You can then group everything together by right clicking on one of the selected items, and choosing to grouping from the menu. At this point you can right click on the group and select save as picture from the menu, then change the file type to JPEG the default isn't a valid file type in the forums. at this point you can place the picture in your post and by learning how to do this you will get practice making diagrams that you could use in reports. I am including an example of a file I've used before.

Are you saying that the field is in the X-Y direction and the pole sticks up in the Z direction with the angles being from the corners of the rectangle to the top of the pole?

The quick reply to ths thread is that weight is the force a palnet exerts on a mass. it has the same units as force because they are the same, however not all forces are weights. If I spin a mass on the end of a string the string has a certain tension this is not the weight of the mass.

for the first one the ratio of the top left side to the top right side is the same for both triangles, so: [math]16/12 = 12/x[/math] [math]16x = 144[/math] [math]x = 9[/math] same idea for y: [math]12/15 = 4/y[/math] [math]12y = 60[/math] [math]y = 5[/math] for the second one triangles ACD and CBD are similar to ABC so the ratios of sides are the same like the previous problem.

the key to this problem is to write what you know. [math]Fg(0)=144 lbs[/math] [math]dh/dt = 3 ft/min[/math] this is constant so the eqtn for height is: [math]h={3 ft/min} * t[/math] if half the weight is gone when the bag has been lifted 18 ft then [math]t={18 ft}*{1 min/3 ft}[/math] [math]t=6 min[/math] so we know know that: [math]Fg(6 min) = 1/2*144 lbs = 72 lbs[/math] so we can determine from this that the eqtn for Fg is: [math]Fg(t) = 144 lbs - (12 lbs/min)t[/math] recall that [math]W=\int_a^b F dh[/math] our particular problem has a hitch in that Fg is a function of time and we need to integrate with respect to height, h. We can fix this though because we have h as a function of time. [math]h={3 ft/min} * t[/math] we can solve for t [math]t=h*{1min/3 ft}[/math] we can plug this into Fg to make it a function of h [math]Fg(h) = 144 lbs - (12 lbs/min)(h*1min/3 ft)[/math] [math]Fg(h) = 144 lbs - (4 lbs/ft)h[/math] now we can integrate this [math]W=\int_a^b F dh[/math] [math]W=\int_0^18 144 lbs - (4 lbs/ft)*h dh[/math] [math]W=144lbs* h -(2 lbs/ft)h^2 |_0^18[/math] [math]W=2592 ft lbs - 648 ft lbs = 1944 ft lbs[/math] or 2.5 BTUs

If we assume that the average density of raindrops is constant than we can calculate the how much strikes the object. We start by looking at a unit time of 1 second, In the picture the height h is the distance the rain falls in 1 second and the distance d is how far the object travels in the same second. The wire frame shape represents the volume of rain which strikes the object. The volume that strike the object is: [math]Vfront + Vtop =(Afront + Atop){\sqrt{h^2 + d^2}}[/math] then we can substitute for h and d [math]h=Vrain*t[/math] [math]d=Vmass*t[/math] [math]Vfront + Vtop =(Afront + Atop){\sqrt{(Vrain*t)^2 + (Vmass*t)^2}}[/math] We could use the ratio of the two velocities to determine an angle for the volume of rain striking the object. Then we could use the equation of the volume as a function of that angle to figure out when the Volume is maximized. If the rain is at terminal velocity than it would only depend on the velocity of the mass.

For the first problem you can use the fact that the two triangles share an angle and that means the ratio of two sides of the triangle are equal for any triangle for the second one use the properties of angles and perpendicular lines. That should help but I don't think I should give you the answer. I would however like to see your conclusion when you get it.

you need to use an infinite series to find it.

If you replaced the Sun and Earth with two black holes a short distance to each side of you, you can sure bet you'd feel yourself being pulled apart... If you replaced the Earth and the Sun with black holes so that you were still at a langrange point than you would not be pulled apart by them.

Personnally I think that everything should be looked at in terms of energy. Rest energy, kenetic, potential. I think that knowing all the energies associated witha particle one could write a total energy distribution for a particle as well as a field equation for the movement of energy in a system. I know that it would make some physics more difficult but it might make unification easier.

I think a better way to say it is that we look at all velocities as fractions of c.

This sounds like you are trying to slip to light speed by a technicality. My first question to you is have you read Einstein's book on relativity? Second have you taken any classes on relativity I only ask because this concept is discussed in basic physics. The idea is that when you are in an inertial reference frame then you appear to be stationary to yourself. That means no acceleration so no rotation, since rotation is constant acceleration around a point. but witha large enough merry-go-round we would not perceive the rotation, just like on Earth. Anyway, in our reference frame we don't see the other MGR coming at us at c because of linear contraction and time dilation. They keep the other object below c. The important thing to remember is that in our inertial frame we think we aren't moving so to us a horse on the other MGR is traveling at a fraction of c, and to a rider on that MGR they think the same thing about us.

Do you just have to build the device or do you need to explain how it works and guess how high it might go? It doesn't seem like there is much physics in just building the thing, engineering yes. If I were your teacher you would be explaing the concepts behind how it works and how high you think it will go.

To try to add to this: A charged particle has an electric field associated with that charge. Fields follow superposition rules so at a point in space we can sum the field vectors for all charges. This gives us the vector for the field at this point. A magnetic field is produced by a change in an electric field. The only way for the electric field to change is to move charges around (or add/subtract them from the system) what ever you do you are still moving them. If you want more of an explanation look up Maxwell's equations on the internet.

The thing to remember about about a force is that it can be defined as a change in momentum and kinetic energy can be defined as the integral of a force. with this logic we could define force a a derivative of both the momentum and the kinetic energy. The difference between the two comes from the fact that when we use momentum we are taking the derivative of momentum with resect to TIME while in the case of the kinetic energy we are taking the derivative with respect to velocity.

When I took undergrad Physics the cat was used to explain why when we pass light through a double slit we see a diffraction pattern, but if we place a detector at one of the slits to tell us when the particle passes through that particular slit we get a pattern as if there is only one slit. The reason for this is because as long as we give the particle a "choice" of paths then there is a probablity that the particle will pass through each slit, then on to the screen. When we place a detector there then we change the system, this change comes from the fact that we can not detect a particle without causing some change in the particle. and this is the essence of the cat experiment. To recap, there is a probability associated with the cat being alive as well as one with the cat being dead. As soon as we open the box to look inside we see which one it is. I hope my explanation is sufficient.

I am talking about a mathematical equation, probably a differential equation of some kind, which tells me how long it takes and I don't mean some general high school physics equation.

I understand that atmospheric pressure and the potential energy of the fluid are general what cause the siphoning effect. The reason I ask the question is that during my research project during my undergrad I used a siphon to move water to study other aspects of fluid dynamics. During this time I was never able to find a complete mathematical explanation of what is going on during siphoning.

I understand the idea behind a siphon, however I have never found any mathematical explaination that states how long it will take for water to flow. I've seen fluid dynamics text which talk about fluids flowing from vessels but never specifically about siphons. If anyone has any direction then I would appreciate it.