03-19-2014

Electric Field

Activphysic:

• For #1 both molecules's charges are positive means each molecules has force points away from the center , so closer means bigger repel force.
• For # 2, first solve the force by F = kq1q2 / r^2(distance between), and solve electric field by E = F12 / q2^2
• For # 3, it basically talk about what electric field looks like on positive charge and negative charge.
• For #4, uniform means the charge between one plates with 10 positive charges and one plates with 10 negative charges charges are always the same.

• For #5, charges in uniform spaces doesn't depend on location because they are all the same.

• For #6, in the uniform space force also stay constant.

Superposition of electric field vector

Can be easily solve by E = F12/ q2^2

E-Fie;d Vectors from a Uniformly Charged Rod

Since the distance is different on each points so we list all the charges by Excel.

We determine the total electric field for x is 5.965*10^5 N/C.

Total electric field for y is 1.273*10^6 N/C.

Games:

 It is impossible for only one charges

A little bit harder, but same trick as first one.

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Just the Practice of Electric field vector

03-17-2014

Exploring the Nature of Electrical Interactions:

The (+) and (-) charges move between conductors, that prefer the sticky side of the taps have same charge due to exchange charge between the table then the non-sticky side and sticky side will attract each other.

• 2 strips attract with different sides such as sticky to non-sticky side.
• The smaller distance between strips give stronger repel force or attract force.
• The 2nd strips tape on the 1st strip follow the same situation; therefore, top to top repel due to the same charge, bottom and bottom repel due to the same charge, and  top to bottom attract due the different charge.

Electric Force Law Video Analysis Activity

Electrical Force vs Separation Distance

Due to the free body gram, Fx and Fy should be zero due to the equilibrium. And it comes out with  F=mg*tan(data), since right triangle y(heigh) - (L^2 - d^2)^1/2, tan(data) is = d /(L^2 -d^2)^1/2. The result is mg*d/(L^2-d^2)^1/2.

The electrical force law is F = kq1q2 / r^2.

• We can calculate the charge on the balls when they have same charges that F = kq^2/r^2
• With q1=1/2q2, we can caculate q1 and q2 with F=(1/2)kq2q1/r^2.
• It is possible to determine the sign of charge on either ball. Since they repel each other, they must have same charges , but not sure is (+) or (-).
• The uncertainty is about 13.2%, the data didn't fit the curve by the external obstruct like air may exchange charge with object, gravity force affect (even it is vary small), and the resist over the rope.

03/12/2014

Extra Credit

How it works:

03/10/2014

Analyzing the Cycle:

Here comes the detail of whole cycle:

• The work has only be done between point 1 to 2 and 3 to 4 because the change of volume is similar to displacement since W=F*d.
• During the step 2 to 3 and 3 to 4, the heat energy transferred to the gas from a reservoir because delta E(int) = Q - W, 2 -> 3 delta E = -3450 J = Q - W ,but W is 0, so -3450 J = Q also from step 3 to 4 :delta E(int) = Q - W, 3->4 delta E =-2370 J = Q - W, and W = -1580 (P*delta V) -2370 - 1580 = Q = -3950 .
• During the step 1 to 2 and 4 to 1, the heat energy transferred to a reservoir from the gas because delta E(int) = Q -W ,and since the Q is positive after calculation the heat is coming into the reservoir.
• The work done is part A the cycle is positive by W = P*delta V which change V is positive cause the W become positive. However, the part C is negative since its change V is negative.
• The net work is equal to sum of work which 2040 + (-1580) = 460 J.

No. 6

Q=mCpdeltaT

Cp= Q/mdeltaT

Cp = 102 J / (1 mole * 5 K ) 

Cp = 20.4 J / mole

No. 7

Cp = 12465 J / (1 mole * 600 K )

Cp= 20.8 J / mole

No . 8

Here comes the way to derive the constant pressure.

Carnot Cycle:

First we solve E on each points, and get the changing E by subtracting two points.
A to B and C to D are both isothermal expansion and compression.
B to C and D to A are both adiabatic expansion and compression.

For isothermal, we use W = P1V1ln(V2/V1) to solve the work then since Q=W in isothermal , we don't need to solve Q again. Of course the delta U  is zero.

For adiabatic, we use W = (P1V1-P2V2)/(r-1) to solve work, and Q = 0 in adiabatic. So the delta U = Q -W  => delta U = -W

03/05/2014

Final Temp Inside the Tube:

Here comes the items, a seal glass tube which can squeeze the air to a small volume and increase the pressure inside the tube to make the temperature rise enough to burn the cotton.

Since smaller volume with same amount of molecules gives higher pressure, the higher pressure cause the higher temperature. Pretend we do this lab pretty fast that almost 0 heat is losing. This means we can use the relationship between temperature and value to calculate the final temperature. The final temperature is 1565.6 degree F.

Activphysic 6 Questions:

Here are six question which involve Isobaric, Isothermal ,and  Isochoric.

Calculation:

Isobaric means constant pressure

Isothermal means constant temperature

Isochoric means constant volume

Using Ideal Gas Law:

Since Energy (work or heat) remand the same at initial and final, the final value and pressure can be easily solve when there are only one unknown.

Since Energy (work or heat) remand the same at initial and final, the final value and pressure can be easily solve when there are only one unknown.

03/03/2014

Volume as Function of Temperature:

Heat up the can with water inside, and put it into the water with room temperature. As result, the can rapidly implode because when steam inside the can immediately condenses the inside can become vacuum for a split second. When outside pressure bigger than inside, the can crash by the outside pressure.

In this Lab, we testing the syringe volume change on different temperature. The hot temperature should increase the syringe volume, and cold temperature should decrease the syringe.

Here comes the testing result, and we calculate the flask volume by V= mass/density.

 This is the lab graph shows volume v.s. temperature, and it clearly shows a positive relationship between volume and temperature. The slope is 0.3020 c.c / *C (* = degree)

• V=(nR/P) T, nR/P is just a constant.

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Two previews Lab data:

This is Pressure v.s. Temperature graph, and the slop is 0.2259 kPa / *C.

This is Pressure v.s. Volume graph, and the slop is 32.88 kPa/c.c.

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According to the comparison of formula and data, we know P=C1T is Pressure vs Temperature graph, P=C2T is Volume vs Temperature graph, and P=C3/V is Pressure vs Volume graph. Therefore, C1 = 0.2259 (kPa/*C), C2 = 0.3020 (c.c/*C), and C3 = 32.88 (kPa/c.c). C3 is come from P=C3/V and since it is a downward slop V is up side down, so C3 is just the slope of graph. Next, C4 = (C1 + C2 + C3)^1/2 which can be use one the formula PV/T=C4. As result the coefficient (C4) is 5.77 (unit confusing, guess kPa/*C + c.c./*C+kPa/c.c.)

Testing the difference of balloon volume with different pressure. This one is the normal room pressure.

When pressure decrease the balloon volume increase because balloon's own pressure is bigger than the outsider pressure.

Picture missing

When pressure increase the balloon volume decrease because balloon's own pressure is smaller than the outside pressure. Additionally, the balloon will goes back to its original size after pressure back to normal.

 Testing the difference of marshmallow volume with different pressure. This one is the normal room pressure.

When pressure decrease the marshmallow volume increase because marshmallow's own pressure is bigger than the outsider pressure.

When pressure increase the marshmallow volume decrease because marshmallow's own pressure is smaller than the outside pressure. But the marshmallow won't back to its original size after pressure back to normal.

02/26/2014

Thermal Expansion:

Testing the ring and stick's different after heat up.

Linear Expansion by using angle data to determine it.

Using steam to heat up the stick, and use angle data
to calculate the expanded length.

object expand when temperature up, and squeeze when temperature down  

Using the angle to solve the circumference to solve the actual length expansion.

------------------------------------------------------------------------------------------------------------Two ways of calculating uncertainty

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Latent Heat:

 Heat up the water~!!

The temperature vs Time graph is a linear graph
which means the Heat Capacity for water is
constant.

The horizontal line shows that water reach
100*C

It will keep a while and go for another linear
graph again if we give water enough energy

Same Lab but by ourselves

The temperature vs Time graph is a linear graph
which means the Heat Capacity for water is
constant.

The horizontal line shows that water reach
100*C

It will keep a while and go for another linear
graph again if we give water enough energy

Due to the Q has already given, latent heat can be solved, and as we known, it's a constant number.

Different because the water spill during the lab, and
also some of heat lose when it changes to steam
which means it become the energy to support the
Latent heat process.

Practice of Qi = Qf

If phase change, the Latent Heat should involved











Ideal Gas:

Pressure means the mass above you that push
down by gravity.

If is liquid then it's the cylinder volume over object, and decided by its area.

Pressure vs Area

It's a parabola slop

We can see that same force gives smaller area more
pressure than bigger area.

It's not a linear means that it not a constant change

Testing pressure on different temperature.

On the pressure vs temperature graph is a linear slop.

The pressure increase when temperature increase, and decrease when temperature decrease.

This means temperature change constantly with delta T.