EET Learning Statement for Outcomes - Develop and Prepare

h = 20cm = .2 m

Vf = 10/.1739 Vf = 57.5

PEi = (.02)(9.8)(.2) PEi = .04

KEf = ½(.02+.2)57.5² KEf = 363.69

Error Analysis: .04 – 363.69

181.87 x 100 = 200 % difference

Experiment Two:

m (ball) = .28kg

t = .55s

h = 1.47 m

Vf = 2(1.47/.55) Vf = 5.35

PEi = (.28)(9.8)(1.47) PEi = 4.03

KEf = ½(.28)(5.35²) KEf = 4

Error Analysis: 4.03 – 4.0

4 .015 x 100 = .74% difference

Experiment Three:

m (myself) = 91kg

t = 4.65s

h = 3.58 m

P = (91)(9.8)(3.58)/ 4.65s P = 686.59W

Php = 686.59W / (746W/h.p.) Php = .92 h.p.

In Conslusion, the difference in potential from an object that is sitting horizontal then pulled by a hanging object ( experiment one) is very large compared to the potential difference in a object free hanging then droped where the potential energy and kinetic energy is reletively the same. The third experement displays the need to enhance my exersise program calculating the horsepower expended during the run was .92 h.p..

Tab 7-C

Troy Szatkowski Physics II 2012-06-PHY-112-OL009

Assignment 5

1. How many calories are needed to raise 500 grams of water from 25 degrees Celsius to 40 degrees Celsius?

The number of calories that raise the temperature of 1g of water is 1 calorie. The temperature change is =15°C. That would mean that 500gx15cal/g would equal 7500 calories.

2. Define the following terms: a. Internal energy – The total of all molecular energies, kinetic plus potential, that are internal to a substance. b. Specific heat – The quantity of heat per unit of mass required to raise the temperature of a substance by 1 Celsius degree. c. Entropy- A measure of a disorder of a system. d. Temperature- a measure of the average translational kinetic energy per molecule in a substance, measured in degrees Celsius or Fahrenheit or in kelvins.

3. Suppose a 1 meter long bar expands 0.5 centimeters when heated. By how much will a 100 meter long bar of the same material expand when similarly heated?

The bar would expand to 50cm. A 1 m long bar expand to .5cm then when heated. The extended bar of 100m would expand .5cm for every meter therefore expanding a length of 50 cm when heated similarly.

4. State how much energy is transferred in each of the following cases:

To melt 1g of ice requires 80cal/g or 335 joules. To vaporize 1g of water requires 540 cal/g or 2255joules a. 2 grams of steam at 100 degrees Celsius condenses to water at 100 degrees Celsius.

2gx540cal/g = 1080calories or 2gx2255joules/g = 4510joules b. 2 grams of boiling water at 100 degrees Celsius cools to ice water at 0 degrees Celsius.

80calx100 = 8,000cal per g x 2g = 16,000calories or

335joulesx100 = 33.5 k joules per g x 2g =67 k joules c. 2 grams of ice water at 0 degrees Celsius freezes to ice at 0 degrees Celsius.

80cal/g x 2g = 160calories or 335joules/g x 2g = 670 joules d. 2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius.

Steam Water - 2gx540cal/g = 1080calories or 2gx2255joules/g = 4510joules

Water 100°C Water 0°C - 80calx100 = 8,000cal per g x 2g = 16,000calories or

335joulesx100 = 33.5 k joules per g x 2g =67 k joules

Water 0°C Ice 0°C - 80cal/g x 2g = 160calories or 335joules/g x 2g = 670 joules

16,000cal + 1080cal + 160 = 17,240 calories or

4510 joules + 67,000 joules + 670 joules= 72,180 joules

5. State and explain the first and second laws of thermodynamics.

The first law of thermodynamics states “When heat flows to or from a system, the system gains or loses an amount of energy equal to the amount of heat transferred.” A system can retain a certain amount of energy, for instance a pot of 100°C water, but when the systems begins to transfer that energy or gain energy by means of heat transfer it cannot be more than the amount transferred by heat. The system cannot generate more energy than what is initially in the system by means of transfer and if energy is transferred to the system it increase the internal energy or increases the amount of transferred energy. As you add heat energy to a pot of water the internal energy increases the temperature of the water but the pot of water, as a system, cannot create the energy internally to increase its own temperature.

The second law of thermodynamics states “Heat of itself never flows from cold object to a hot object.” This law seems to be very straight forward and easy to translate. A cold object cannot transfer energy to a hot object.

6. Calculate the change in internal energy for each of the following systems: a. 100 joules of energy is added to a system that does 40 joules of external work. If 40 joules of external work is used then 60 joules of internal energy remains.

b. 75 joules of energy is removed from a system that has 50 joules of work done on it.

The internal energy need to be at least 150joules. 75+50 = 150 joules c. 150 joules of energy is added to a system that has 75 joules of work done on it.

If 75 joules of work is used then 50 joules of internal energy remains.

7. Explain why a body of water freezes from the top down rather than from the bottom up.

The ice molecules are less dense than the water around the ice. When water freezes to 0°C it becomes less dense. The crystals form in the water then float to the top because they are less dense. If the ice crystals were less dense than a liquid they are immersed in then the ice would sink and the freezing would start at the bottom.

Tab 8-C

Troy Szatkowski

ELEC152: Circuit Theory I

M7A3: Lab: Maximum Power Transfer

Introduction:

This lab demonstrates the maximum Power Transfer. The outcome of the lab gives a perspective on principles characteristics of the Maximum Power Transfer as it applied to a circuit utilizing the mathematical applications and the practical application acquired by recording readings.

Lab Experiments:

1) Setup the following circuit in MultiSim.

2) Measure the open-circuit voltage VAB. This is done by connecting a voltmeter across the A and B outputs. Then measure the short-circuit ISC. This is done by connecting an ammeter across the A and B outputs.

3) VAB, the open-circuit voltage, is the same as the Thevenin voltage. ISC, the short-circuit current, is the same as the Norton current. Dividing VAB by ISC will give the value of the Thevenin resistance RTH (or Norton resistance).

Find the RTH for the circuit and record its value here:

4) Add a fourth resistor to the circuit, across the outputs A and B, with the meters connect to measure its voltage and current.

Simulate the circuit for each resistance in the following table and record the current and voltages or each value.

Conclusion:

The lab went well but with an issues utilizing the excel spreadsheet charting system. The lab required to transpose voltage, current and power on a chart. I could not successfully implement all the recordings on one chart. I improvised by designating each value into three separate charts utilizing the resistor value as the x value. The remaining experiment conducted within the lab when well. I acquired to results for power, voltage and current as expected.

Current

3 K Ohms

Resistance

Current

Power

3 K Ohm

Resistance

Power (mW)

Voltage

3 K Ohms

Resistance

Voltage

R1

1000Ω

R2

4000Ω

R3

2200Ω

R4

100Ω

V1

30 V

XMM1

XMM2

V1

30 V

R1

1000Ω

R3

2200Ω

R2

4000Ω

A

B

V1

30 V

R1

1000Ω

R3

2200Ω

R2

4000Ω

XMM1

XMM2

A

B