AP Physics 2 Unit 6 Lab

Elisa Alvarado, Sarah Cratem, Ryan Partain, Julia Reidy

Mr. Thomas

AP Physics 2 cmod

13 December 2015

Unit 6: Charged Particles Flow Model Lab Report

Objective: To determine the mathematical and graphical relationship between electrical potential and time for capacitors being charged and discharged when connected to resistors of varying resistivity in series.

Apparatus:

Procedures:

1. Set up the circuit using the 22,000 ohm resistor
2. Charge the capacitor
3. Discharge the capacitor
4. Switch the 22,000 ohm resistor with the 47,000 ohm resistor
5. Charge the capacitor
6. Discharge the capacitor
7. Switch the 47,000 ohm resistor with the 100,000 ohm resistor
8. Charge the capacitor
9. Discharge the capacitor

Data Tables: https://drive.google.com/a/bishopkennyhs.org/file/d/0B2kSUAkE0NVYNGRXcENwUDFncVU/view?usp=docslist_api

Graphs:

Potential vs. Time

Conclusion: While the capacitors are charging, the potential and time are related by the following equation: Vc=Vsource(1-e^(-t/RC)). The graphical relationship between V and t for charging and discharging is represented by the equation (1-ln(Vc/Vs))=-t/RC. While the capacitors are charging, the potential and time are related by the following equation: Vc=Vsource(1-e^(-t/RC)) which can be converted to ln(1-(Vc/Vs))=-t/RC.  We plotted Vc/Vs for the purpose of finding the slope of the final equation and linearizing the graph. The slope was found by the equation slope=1-ln(Vc/Vs)/r where r=resistance with the units of ohms/s. The resistance of the resistor affects the system by lowering the potential and increasing the time it took the circuit to charge. As the capacitance went up the charge and stored charge also rose. Some error could have occurred through a misreading of the devices, batteries not being completely charged, or small amounts of internal resistance on the circuit. The error could be minimized by making sure all equipment is up to date and read properly and carefully.

AP Physics 2 Unit 4 Lab

Elisa Alvarado, Sarah Cratem, Ryan Partain, Julia Reidy

Mr. Thomas

AP Physics 2 cmod

13 December 2015

Unit 4: Charged Particle Interaction Model Lab Report

Objective: To determine the mathematical and graphical relationship between electrostatic force and distance between a neutral and charged object, and to determine the graphical and mathematical relationship between electric force, distance between a neutral particle and a charged particle (r), and the distance traveled by the neutral particle (x).

Apparatus:

Trig Relationships and Force Diagram:

Procedures:

1. Place physics stand 1 meter apart and place the meter stick on top of the stands. Tape the meter stick to the stands. Apply a charge to one balloon by rubbing it against your hair.
2. Tie the strings to the stands and tie the one balloon to each string
3. Separate the charged balloon from the uncharged one until the uncharged one is unaffected by the charge and is at rest directly between the stands.
4. Bring the charged balloon closer to the neutral balloon until the neutral balloon moves and measure the distance between the two and the distance the uncharged balloon was displaced.
5. Repeat the 4th step 8 times and be sure the same increments are used for each trial. Record the space between each of the balloons and how much each balloon is displaced during each trial.
6. Calculate the electric force using net forces and trigonometry as shown in the diagrams above.

Data Tables:

Graphs:

This shows an inverse relationship between delta x and r. To linearize this graph we must take the inverse squared of the quantity r.

displacement vs. 1/r^2

This graph shows a linear relationship between displacement and 1/r^2, as represented by the equation Δx = (1.53 x 10-4m3)/r2.

This graph shows an inverse relationship between electrostatic force and distance between the two balloons. To linearize this relationship we must take the inverse squared of the quantity r.

This graph demonstrates a linear relationship between electrostatic force and the inverse squared of r, represented by the equation Fe= (7.768x10^-6 Nm^2)(1/r^2).

Conclusion:

Our first graph showed an inversely proportional relationship between the change in position of balloon 2 and the distance between the two balloons. To linear use this graph we took the inverse quaked of the quality r. This left us with the relationship of delta x is proportional to 1/r^2.  Since electric force follows this same relationship, we can infer that Fe is proportional to 1/r^2. To determine the electric force we used the force diagram above. From this force diagram we used trigonometry to find the equation Fe = mgtanθ. When you solve for theta using trigonometry the new finalized equation comes out to be Fe=mgtan(Δx/0.2). M is the mass and g is the gravitational constant. To make a general equation to relate the electrostatic force to 1/r^2 we added in a constant k to represent Coulomb's constant. So that Fe = k/r^2. Since the force and the charges show the same proportional relationship to one another we added in Fe=(kq1q2)/r^2. The slope from out linearized graph gave us the number 7.7e-6Nm^2  which represents the product of Coulomb's constant and the charges. ( 7.7e-6Nm^2 =kq1q2)

Error in this lab could have resulted from human errors from measuring the exact distances the objects moved. There could have also been error with the charge of the balloon being greater or lesser at some time.

AP Physics 2 Unit 3 Lab

Elisa, Sarah, Ryan, Julia

Mr. Thomas

AP Physics 2 cmod

15 September 2015

Unit 3: System of Flowing Particles Model Lab Report

Objective: To determine the effect of volume of a tube on distance and volume of water expelled through different-sized holes.

Apparatus:

Procedures:

1. Connect sections of tube to a height of 153 cm from bottom of tube up to the excess hole.
2. Turn hose on and place in the open top of the tube. Wait for the water to rise to and exit out of the excess tube.
3. Measure the horizontal distance the water exits the horizontal holes in the tube from the base of the tube to the end of each stream.
4. Measure the volume of water expelled from each hole in [] seconds by holding a bucket in each stream for the set time.
5. Repeat this procedure for tube heights of 142 cm, 119 cm, and 57 cm, and when measuring the volume of water expelled in a given time, for [] seconds, [] seconds, and [] seconds, respectively.

Data Tables:

Graphs:

Velocity vs. Height

The relationship between velocity and height creates a curve. To linearize this, the values for velocity must be squared.

Velocity Squared vs. Height

The relationship between velocity squared and height is proportional, described in the equation v^2=4 m^2/(s^2 cm)h. This relationship should, however, be described in the equation v^2= (2g N/m)h, where 2g is the slope.

Flow Rate vs. Height

The graph of flow rate vs. height has a relationship that results in a curve. To linearize that we must square the flow rate.

Flow Rate^2 vs. Height

The graph of flow rate^2 vs. height shows a proportional relationship as described in the equation Iv^2=(121 m^5/s) h.

Flow Rate vs. Velocity

The relationship between flow rate and velocity is proportional, as described in the equation Iv=(5.67 m^2)

Flow Rate vs. Area

The relationship between flow rate and area, when the height of the pipe is 153cm, is linear, as shown by the equation: Iv=2.773E-04 m/s(A) - 3.243E-06 m^3/s.  The slope of this line is velocity.

Velocity vs. Area

As shows by the data, the exit velocity of the water is not affected by the area of the hole.  All of these values were obtained when the height of the pipe was 153cm.

Conclusion:

The relationship between velocity and the height of the pipe is that the higher the height, the greater the velocity becomes. An equation that can be used to describe this is  v^2=2gh. The relationship is between flow rate and velocity is that they are directly related and it follows the equation, I=AV. These graphs show that the velocity of a fluid is dependent on the height of the tube and the area which it travels through. The flow rate depends on the area of the hole that it flows out of. The slope of the velocity squared vs. height graph shows us that the slope should be 2 times the gravitational constant "g". The slope of the flow rate vs. velocity means that the velocity and flow rate are proportional to one another. There are no y intercepts for the graphs because they are proportional. Some errors in this experiment could have been with our trials because we didn't have enough time to do all of the trials we wanted to.  Also, errors could have happened with our measurement of the distance the water sprayed to find the velocity, because it was very hard to tell exactly where the water hit the measuring tape. Another error that could have happened was with the angle we held the pipe up with, it could have been not completely straight up. It is also possible that small amounts of water could have been lost while transferring it from the containers to the beakers or graduated cylinders for measurement. These possibilities may have caused varied results in the experiment.

AP Physics 2 Unit 2 Lab

Elisa Alvarado, Sarah Cratem, Ryan Partain, Julia Reidy

Mr. Thomas

AP Physics 2 cmod

9 September 2015

Unit 2: System of Ideal Particles Lab Report

Objective: To determine the effect of pressure of a system on number of particles, volume, and temperature of the system.

Apparatus:

Procedures:

Pressure vs. Volume:

1. Hook up a syringe to a pressure sensor and take a flask with a stopper attached to it. Attach one hole to the pressure sensor and keep the other end open to be able to open and close it.
2. Fill the syringe up halfway and fill the container and inject the syringe directly into the container.

Pressure vs. Number of Particles:

1. Fill the container up to 20 mL.
2. Open the valve.
3. Fill up syringe halfway. Empty the syringe into the container.
4. Close the valve.
5. Repeat steps 2-4.

Pressure vs. Temperature:

1. Start with close to boiling water.
2. Place the flask and thermometer in a water bath, making sure the water is close to boiling point.
3. Put the magnetic stirrer at bottom of water bath.
4. Submerge the flask in water with the magnet spinning and leave it until it hits equilibrium.
5. Measure the temperature of the water bath.
6. Connect a single-hole stopper to a pressure sensor and measure the pressure.
7. Place small handfuls of ice into water bath and wait until it hits equilibrium.
8. Measure the temperature of the water bath.
9. Repeat steps 2-8.

Data Tables:

Graphs:

Number=(5.577kPa/#)(pressure)-(569.017)

Temperature=(4.681kPa/°C)(pressure)-(401.957°C)

Volume=(875.922mL/kPa)/pressure

Conclusion:

Pressure and volume are inversely related and their relationship can be described in this experiment with the equation Volume=(875.922mL/kPa)/pressure. This forms an inverse graph. Temperature and pressure are directly related and their relationship in this experiment can be described with the equation Temperature=(4.681kPa/°C)(pressure)-(401.957°C). This creates a positive linear graph. Number and pressure are also directly related and can be described with the equation Number=(5.577kPa/#)(pressure)-(569.017) in this experiment.  This also creates a positive linear graph. All of the relationships can be described together with the equation PV=nRT, where volume is measured in liters, pressure is measured in atmospheres, n is the number of moles, the temperature is measured in Kelvin, and R is a constant.  This equation can also be written in another form to work with different units (Pascals for pressure, cubic meters for volume, particles rather than moles, and a different constant, k): PV=NkT.

AP Physics Lab 11

Electrically Charged Particle Model Lab

Brooke Miller, Gabrielle Murphy, Katie O'Byrne, Ryan Partain

Objective: To find the graphical and mathematical relationship between potential and current in a light bulb and two different colored resistors.

Apparatus

Materials:

Battery
Lightbulb
Switch
Dimmer Switch
Resistors (Blue and Green)

Multimeter

Procedure:

1. Create a circuit on the board with the battery, switch, dimmer switch, and bulb connected by wires. Put the dimmer switch on the highest setting.

2. Open the switch to break the circuit. Measure the current across the open switch using the multimeter.
3. Close the switch to close the circuit.
4. Measure the potential drop across the light bulb using the multimeter.
5. Turn the dial on the dimmer switch to the second highest setting and repeat steps 2-4.
6. Repeat steps 2-4 with all the spots on the dial.
7. Take out the bulb and replace it with the green wire (resistor).
8. Repeat steps 2-6 with the green wire.
9. Take out the green wire and replace it with the blue wire (resistor).

10. Repeat steps 2-6 with the blue wire.

Data

Circuit with Light Bulb

Circuit with Green Resistor

Circuit with Blue Resistor

Data Analysis

Lightbulb

Lightbulb Linearized

V=(49.71 V/A2) I2

Blue Wire

V= (9.837 V/A) I

Green Wire

V= (4.937 V/A) I

Conclusion

With the light bulb in the circuit, the potential vs. current graph formed produces a parabolic shape. In order to linearize the data we squared the current. The line gave us the equation V= (49.71 V/A2) I2 where V is potential drop, I is current and the y-intercept is where the current equals zero but is close enough to zero that it is negligible. The slope of the graph represents how difficult it is for the current to move through the circuit, or the resistance present within the circuit.

With the different resistors in the circuit, the potential versus current graphs produce a straight line. As with the experiment with the light bulb, the slope of the line represents the resistance of the wire. The difference between the green and blue resistors were their slopes, or the resistance within the  system; the green wire had a resistance of 5 ohms or (V/A) and the blue wire had a resistance of 10 ohms. This gives us the equations V=(4.937 V/A) I and V= (9.837 V/A) I respectively.

The general equation found for both the blue and green wires is V=IR, where V is the potential drop, I is the current, and R, the slope, is the resistance.

The blue and green resistors have a linear relationship between voltage and current. The graph of the voltage vs current in the light bulb forms a parabolic shape with a changing slope. This difference between the slopes of the green and blue wires and a light bulb because a light bulb is not a “fixed resistor.” While the currents flowing through the systems in each trial are similar throughout all three experiments and the battery is the same, the resistance changes in each trial of each experiment depending on the dimmer switch, which adds resistance. When fixed resistors such as the wires are part of the circuit, the resistance of the wires don’t change even when resistance is changed with the dimmer switch.

Some sources of error could have been that the resistor did not function as well as planned and it had a larger percent error than predicted. The multimeter used also might have been slightly inaccurate and measured incorrect numbers. Energy might have been taken out of the system through heat. The voltage is also slightly lower than expected because potential is lost in the movement across the wires and as the battery continues to power the circuit.