s pepperchin

I would say that people should have a basic understanding of science, however I think that as people who are scientist it seems kind of elitist to tell people they need to know organic chemistry, nuclear physics, cosmology or any specialized area of science. I don't know enough about art to be a master painter and I don't expect a master painter to know enough about physics to do quantum mechanics.

I checked one of my physics books and the equation for the Potential energy of an inductor is:

[math]U_L=\frac{1}{2}LI^2[/math]

and the equation for inductance is :

[math]L=\mu_0 A ln^2[/math]

when the inductor has a material in the coil the [math]\mu_0[/math] is replaced by

[math]\mu=\mu_0(1-\chi_m)[/math]

[math]\chi_m[/math] is the magnetic susceptibilty of the materieal

However an inductors kinetic energy would just be the relativistic kinetic energy of the electrons in the coil.

This is the equation for reletivistic energy.

[math]E=\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}[/math]

I haven't put these all together yet but I figured that since you were interested I would give you a few eqtns to work with while I also work on it and see what we get.

Hmmm I wonder whether I've answered my own question here. If i is constant then the kinetic energy of the electrons will be constant, but you can add materials into the coil to change the inductance without changing the coil itself.

How does this change in inductance change the kinetic energy of the electricity on the other coil, or the whole system for that matter?

When I worked through it the first time I got the same thing you did but when I looked at the steps the units didn't work out the whole way through. I wasn't sur ehow exactly I had messed up but I plugged the integral into Mathematica and it gave me a weird answer that didn't even seem similar to what I had imput. so at that point I loked at a couple of different factors for the same integral and did some backwards engineering from the actual eqtn.

It sounds interesting but if they want to prove that it works they could take a non-magnetic mass on a spring and create enough gravity to stretch it farther than it is just from the Earth's gravity.

A good example is if you start with an initial momentum of 0 and then shoot two objects (same mass) off in opposite directions (same velocity) then the Conservation of Momentum is still kept as the momentum of each of the two bodies cancel each other out.

 

If you took the magnitudes of each momentum then it would be a non-zero value and so momentum would not have conserved' date=' therefore you cannot do that (take the magnitude of the momentum).[/quote']

 

that is why no one cares about the magnitude of the momentum vector.

 

and going back to the comment about energy as a vector it is not a vector. When you define energy you define the energy at a point in space at a moment in time. If it is changing then you can calculate the energy flux and that would be a vector.

I have been working on an example for using different reference frames.

Say we have a star located at (0,0) in reference frame A and it is orbited by two planets M and N. The position vectors for the two planets are:

[math]\overrightarrow{r}_M=<a Cos(\omega_M t),b Sin(\omega_M t)>[/math]

[math]\overrightarrow{r}_N=<c Cos(\omega_N t),d Sin(\omega_N t)>[/math]

where [math]\omega_M[/math] and [math]\omega_N[/math] are the angular velocities of the planets and the values of a,b,c and d are the semimajor and semiminor axis of the orbits. This means that the three objects have the position vectors:

[math]\overrightarrow{r}_S=<0,0>[/math]

[math]\overrightarrow{r}_M=<a Cos(\omega_M t),b Sin(\omega_M t)>[/math]

[math]\overrightarrow{r}_N=<c Cos(\omega_N t),d Sin(\omega_N t)>[/math]

I decide that I am going to use a new reference frame B which has planet M at the center. I will start by adding a vector to the orbit of planet M such that the position of planet M is always <0,0>.

[math]\overrightarrow{r'}_M[/math]=[math]<a Cos(\omega_M t),b Sin(\omega_M t)>[/math]+[math]<-a Cos(\omega_M t),-b Sin(\omega_M t)>[/math]

when I combine these vectors into one vector I get:

[math]\overrightarrow{r'}_M[/math]=[math]<a Cos(\omega_M t)-a Cos(\omega_M t),b Sin(\omega_M t)-b Sin(\omega_M t)>[/math]

or

[math]\overrightarrow{r'}_M[/math]=[math]<0,0>[/math]

now to determine the position vectors of the other objects I just add the same vector to their position vectors as well.

[math]\overrightarrow{r'}_S[/math]=[math]<0,0>+<-a Cos(\omega_M t),-b Sin(\omega_M t)>[/math]

[math]\overrightarrow{r'}_S[/math]=[math]<0-a Cos(\omega_M t),0-b Sin(\omega_M t)>[/math]

[math]\overrightarrow{r'}_S[/math]=[math]<-a Cos(\omega_M t),-b Sin(\omega_M t)>[/math]

[math]\overrightarrow{r'}_N=[/math][math]<c Cos(\omega_N t)-a Cos(\omega_M t),d Sin(\omega_N t)>[/math]+[math]<-a Cos(\omega_M t),-b Sin(\omega_M t)-b Sin(\omega_M t)>[/math]

So now in reference frame B the position vectors are:

[math]\overrightarrow{r'}_S[/math]=[math]<-a Cos(\omega_M t),-b Sin(\omega_M t)>[/math]

[math]\overrightarrow{r'}_M[/math]=[math]<0,0>[/math]

[math]\overrightarrow{r'}_N=[/math][math]<c Cos(\omega_N t)-a Cos(\omega_M t),d Sin(\omega_N t)>[/math]+[math]<-a Cos(\omega_M t),-b Sin(\omega_M t)-b Sin(\omega_M t)>[/math]

as you can see they are both correct and all physics works in both frames the math is just easier in frame A.

This is how you derive that eqtn:

start with your given eqtns

1 [math]V=V_L + V_R[/math]

2 [math]I_{max}=\frac{V_0}{R}[/math]

3 [math]V_L=L\frac{dI}{dt}[/math]

4 [math]V_R=IR[/math]

Starting from eqtn 1 insert eqtns 3 and 4

[math]V=L\frac{dI}{dt} + IR[/math]

the next set of steps is just some algebra to get it in the form we want before we integrate

[math]V-IR=L\frac{dI}{dt}[/math]

[math]V(1-I\frac{R}{V})=L\frac{dI}{dt}[/math]

[math]V(1-I\frac{R}{V}) dt=L dI[/math]

[math]\frac{V}{R}(1-I\frac{R}{V}) dt=\frac{L}{R} dI[/math]

[math]\frac{R}{L}dt=\frac{R}{V(1-I\frac{R}{V})}dI[/math]

[math]\frac{R}{L}\int dt=\frac{R}{V}\int\frac{1}{(1-I\frac{R}{V})}dI[/math]

at this point we can use the follwing integral that is in most integral tables.

[math]\int \frac{1}{a+bu}du=\frac{1}{b}ln|a+bu|+c[/math]

in this case a = 1 and b = -R/V

[math]\frac{Rt}{L}=\frac{R}{V}(\frac{-V}{R}ln(1-I\frac{R}{V}))[/math]

[math]\frac{Rt}{L}=-ln(1-I\frac{R}{V})[/math]

[math]\frac{-Rt}{L}=ln(1-I\frac{R}{V})[/math]

[math]e^{\frac{-Rt}{L}}=1-I\frac{R}{V}[/math]

[math]e^{\frac{-Rt}{L}}-1=-I\frac{R}{V}[/math]

[math]1-e^{\frac{-Rt}{L}}=I\frac{R}{V}[/math]

[math]\frac{V}{R}(1-e^{\frac{-Rt}{L}})=I[/math]

[math]I=\frac{V}{R}(1-e^{\frac{-Rt}{L}})[/math]

[math]I=I_{max}(1-e^{\frac{-Rt}{L}})[/math]

I hope this helps you

since momentum is a vector and is conserved in a system, that would mean that the scalar or absolute value of the momentum as you put it would also be conserved since that is simply the magnitude of the momentum. This however does not mean that all systems with the same magnitude of the momentum are the same.

You are looking at the problem wrong. The answer says that when t is a half integer multiple of Pi, such as Pi times 1/2, 3/2, 5/2 etc, then x is equal to a,

just like the graph shows. Your solution is a function of time not a.

I would let you write a new paper but as your teacher it is his call what to do. The only other suggestion is to have your parents go to school and complain about how the issue is being handled.

I consider myself a scientist because I am driven to gain knowledge and understand it in a way that I can pass that understanding on to others.

And a B.S in Physics for those who require credentials.

Are you asking if there is a relation between the potential energy of a particle in the magnetic field and the kinetic energy of the electrons flowing through the coil? If that is what you are saying then I am sure there is a way to figure out a relation between the current in the wire and the position and charge of the particle.

Plagiarism on any level is stealing someone else's work. But the actual question here is did you use wikipedia as one of your sources? and if you did why didn't you quote it. Most kids in 8th grade (not that I'm making an excuse for you) that plagiarize because they don't realize that they have written something word for word after they recently read it somewhere. The only solution I can suggest for you is to go to your teacher and tell them that you didn't intend to plagiarize and that you would like an opportunity to write a new report on a new subject. I would also talk to the teacher about citing sources so that it doesn't happen again.

I could define all the math involved in the motion of planets with any point as the center. but the heliocentric math is easier than the rest.

The requirements don't say anything about the 4 digit number A being prime, only that 4 of the 5 two digit numbers are prime.

oops I read it pretty fast and mis understood

I am pretty sure that the yellow squares are correct. It says that none of the two digit numbers can start with zero so that means that top row can't have 0 anywhere. And since the bottom 4 digit number can't begin with 0 the left yellow square should be yellow, the yellow square on the right is yellow because the only way for it to be zero is if the number on top would be multiplied by 5 or 10, or end in a 5. All of these possibilities aren't possible since to have a 5 at the end makes the top number non-prime, multipling by 5 requires the top number to end with an even number (again non-prime), and multipling by 10 can't work because that would make the bottom number 5 digits long. Thats all I have to say for now.

I looked at it after I posted and that is true. I haven't figured it out yet though.

Just from looking at it I would say that you might be able to factor it since .2 is the 1/5th power and you have 32 which is 2^5 and you have x^5. Try that and see what you find.

I've been under the impression that gravity at the center of the earth cancels out. A sentient being in a habitable room there' date=' would experience weightlessness.

 

Can anybody assure me if that impression is correct, or explain why not?

Thank you![/quote']

 

That is correct you would be weightless at the center of the Earth.

 

The reason for this is that the weight you feel is the force the earth exerts on you:

 

[math]F_{Earth} = \frac{GMm}{r^2}[/math]

 

although this suggest that as r-> 0 that [math]F_{Earth}[/math] would go to infinity when we look at what is actually going on we see that your center of gravity is at the same point as the Earth's center of gravity, and so neither of you are pulling the other in any direction.

 

The complete explanation for this is that when you look at the world as a perfect sphere and do a little calculus the only mass that is taken into consideration when determining the force of gravity a mass exerts on another mass only the amount of mass that is a distance less than or equal to r is taken into account. It all has to do with Gaussian surfaces.

Do you have a link where we could see this picture?

Okay heres how it goes:

Start with your three reactions

[math]2 C_2 H_2 + 5 O_2 -> 4 CO_2 + 2 H_2 OH = -2600 KJ[/math]

[math]2 C_2 H_6 + 7 O_2 -> 4 CO_2 + 6 H_2 OH = -3120 KJ[/math]

[math]H_2 + \frac{1}{2} O_2 ->H_2 OH = -286 KJ[/math]

Remember that you want everything to be on the correct side of the arrow when you get to the end, like:

[math]C_2 H_2 + 2 H_2 -> C_2 H_6[/math]

Step 1: Divide the first reaction by 2

[math]C_2 H_2 + \frac{5}{2} O_2 -> 2 CO_2 + H_2 OH = -1300 KJ[/math]

The second reaction reaction by 2 and switch the direction(which includes changing the sign for the energy)

[math]2 CO_2 + 3 H_2 OH-> C_2 H_6 + \frac{7}{2} O_2 = 1580 KJ[/math]

and multiply the third reaction by 2

[math]2 H_2 + O_2 -> 2 H_2 OH = -572 KJ[/math]

Now I add the three reactions together to get:

[math]C_2 H_2 + \frac{5}{2} O_2 + 2 CO_2 + 3 H_2 OH + 2 H_2 + O_2[/math] -> [math]2 CO_2 + H_2 OH + C_2 H_6 + \frac{7}{2} O_2 + 2 H_2 OH = -1300 KJ + 1580 KJ -572 KJ[/math]

It can be simplified into:

[math]C_2 H_2 + 2 H_2 ->C_2 H_6 = -292 KJ[/math]

by crossing out the compounds that are the same on both sides of the reaction.

I have a basic understanding of the polar coordinate system. I have some questions that would help me with my hw though.

 

(1) how do I take an equation like 3cos(5 theta) and figure out how big an interval is for one loop. I know that there are 5 loops' date=' i just don't know how to get the interval.

 

(2)i know that when integrating to get area im going to take .5( integral from angle a to angle b of r^2 dtheta)

 

i also know that if im finding the area between two curves i take the difference of r1^2 and r2^2 depending on whic yields a greater value, would i break the intervals up for when one function was greater then the other.

 

any help is appreciated[/quote']

 

For the first question how do you get that there are 5 loops?

 

for the second you would just take the integral of the outer radius and subtract the area for the inner radius and that would give you your answer.

1. A' date=' B and C are points on the circumference of a circle, centre O, and ∠BAC=115°. Calculate the number of degrees in ∠OBC.

 

I drew a diagram but could not come up with anything primarily because ∠BAC and centre O lie on opposite sides of chord BC (draw a quick diagram if you don't know what I mean).

 

2. ABCD is a quadrilateral inscribed in a circle and AB = CD. Prove that AC = BD.

 

This is pretty simple I think, if AB = CD, then BC = DA (do I need to say why?). Then just use Pythag to prove the diagonals are equal.[/quote']

 

For the first problem you need to look at al of the geometric relations you can get from circles as well as triangles.

 

For the second problem you can't assume that bc = da.

 

You should go into more detail about how you are doing the problem. They are both possible.

You can just use either eqtn to solve for one of the variables, then substitute that into the second equation and get an answer for one of them. After that you would just use that answer in either eqtn to get the value for the other variable.