Lab 2

Lab 1

Lab 7

Activ Physics:

oddly enough the computer would not show the forces. but we were able to calculate the results below.

question 1 and 2

Depending on the amount of field lines we can approximate the electric field of a point charge. the more field lines there are the higher its electric field as well as its charge.

The behavior of a negatively charged particle in an continuous charge electric field.

These are some calculations made in excel. Showing a charged particle reacting to the continuous charged rods. The total electric field is calculated using small increments and calculating the elect field at those given spots. Then adding the individual pieces to get a total.

Electric Hockey

In this game we have to use opposing or attracted charges to navigate a small charge to the goal!

Using 2 positives to cure it around the obstacle.

again using 2 positive charges to flank the positively charged puck into the goal, but this time having more of a y=1/x^2 shaped curve.

(pay close attention to the amount of trials. Perhaps the record has been broken!)

Lab 6

Electric Repulsion of Tape

By ripping off 2 strips of tape from the table...

...we can electrically charge each piece, and depending on their charge they will either repel each other (-) and (-) or (+) and (+)...

...or they can attract each other if there is a positive charge in one tape and a negative in the other.

Observations, and explanations of the tape experiment.

We did some theoretical work on a swing example and came up with a relationship for Electric force. the work flow is a bit scatterbrained but the Force equation in the bubble holds the final most minimized function.

Like Charge Repulsion Experiment  

The initial position plots made from the supplied video

A graph of the repulsion force vector in relation to distance between the two charged objects.

The graph made in logger pro shows that the electrical repulsion force is inversely proportional to the square of the distance.

Although the experimental data doesn't show that the force is EXACTLY inversely proportional to the square, it can be assumed that there is some error in the supplied data. There is a 15% difference between the theoretical and experimental derived formulas.

Explaining the shortcomings of the coulombs law as well as some causes of error

Diesel Cycle

 The Diesel Cycle

Here are all the calculations of the cycles. Using mostly the ideal gas law.

Due to the fact that I totally destroyed the original paper I was forced to write the answers on a whiteboard.

All the resluts were fairly logical.

The entropy came out to be zero. I compared my values to others and they had something like 0.0001, which is existant but negligible, becuase the data provided to us only had 2 significant figues. We cannot create more accuracy (more significant figures) with data than what is provided.

Final Results

Lab 5 - Heat Engines and Cycles

A Simple Gas Cycle Of A Heat Engine

Above we have the graph of a simple gas cycle, where half the processes are isobaric (constant pressure) and the other half are isometric (constant volume). This is an ideal model, and i very unrealistic. 

Carnot Cycle

This cycle is a much more realistic cycle of a heat engine. It is used in gasoline burning engines (your car unless you drive a Tesla or a Diesel). I believe that some of the calculations of Work and Heat a inaccurate, so let me explain the faults for your better understanding:

From B->C and D->A the process is adiabatic, which means that there is not heat transfer, thus the Q for both of those processes should be 0 (zero) J. Thus indicating that the W=-dE. From there on out the calculations should fall into place

We used W=nRT*ln(V2/V1) for the Work of an isothermic process, hence from A->B and C->D.

W used W=((PiVi^y)(Vf^(1-y)-Vi^(1-y)))/(1-y), where y=gamma, in a adiabatic process.

The total work should be around 900J. Most the work is done in the power stroke from A->B.

ActivPhysics 8.7 - Heat Capacity

Problem 6:

In this model the pressure was held constant, as the temperature and volume changed.

Using the molar heat capacity formula we derived 20.75J/mole*K

Using two different temperatures which are fairly close to one another. 

Problem 7:

We did the same process as in problem 6, except that our change in temperature was the maximum possible, observing if there will be any difference than the previous problem.

Not much of a difference.. we calculated 20.78 J/mole*K

Problem 8:

Using a combination of formulas (1st law of thermodynamics, Internal energy, ideal gas law) and rearranging them, we calculated that the molar heat capacity is (5/2)R=20.785J/mole*K

This is very close to our calculations from problem 6&7.

Lab 4 - Kinetic Theory and PV Diagrams

 This is a fire syringe. Capable of creating high temperatures within it, by simply compressing the volume rapidly. We will demonstrate that the fire syringe can ignite cotton.

The calculations above are showing that the initial temperature of the inside of the syringe before compression is 296.5K (room temp). Upon compression the temperature skyrockets to 1381.8K which equals >2000 degree Fahrenheit!!

The combustion temperature of paper (cotton) is 451 degrees Fahrenheit, which will be no problem for the syringe.

we used the relationship of ViTi^(3/2)=VfTf^(3/2) to acquire our final temperature.

What a majestic blow by our professor! well done professor, your pupils cheer in excitement :D

The model for six problems from activphysics.com

It was demonstrating the state variables and ideal gas law. 

These problems were to test our knowledge and understanding of the relationship between Temperature, Pressure, and Volume, if one of them is constant.

Question 1-3 show the relationships between volume VS temperature in a isobaric process, pressure VS temperature in an isochroic process, and pressure VS volume in an isothermal process respectively.

Question 4-6 show the conceptual changes of each process. We thought only the theoretical concept was required. After consulting other students, we might believe that specific calculations were needed.

So I will provide them below:

4)  V=(nRT)/P

V=((1mol)*(8.314)*(301.8K))/101000Pa

V=25.09dm^3

5) P=(nRT)/V

P=((1mol)*(8.314)*(300K))/0.020m^3

P=124710Pa

6) P=(nRT)/V

P=((1mol)*(8.314)*(300K))/0.020m^3

P=124710Pa

Lab 3 - T.vs V. (Charles' Law)

In this experiment we wanted to find the relationship between Temperature and Volume. 

Using an extremely rare glass syringe, and 2 containers with hot and cold water. 

After measuring the temperature and volume of the glass flask and syringe we can see that the Volume is directly proportional to the Temperature. 

We plotted the sparse data into excel and had a line fitted to the plot points. Observing the picture it can be seen that this is a pretty good match. 

The slope of the line is what should be the missing constant of our formula relating Temperature and Volume.

PV=nRT

=> V/T=(nR)/P

(nR/P) = .3242

Looking at these calculations, the units of the coefficient (which are a combination of n, R and P) are (m^3)/K.

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P.s: I figured out what the problem was with my blog. I will upload the first 2 entries ASAP!

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