The distance between the forces is given by the Coulombs law through the use of the formula F=kq1q2/r2.0.1newtons = 8.99*109*3.2*10-6*7.7*10-7/r2 R= 555.78 Answer to question 3 • Potential difference between the two plates is equal to velocity which is equal to 6.0*106m/s • Force = mass *acceleration = 1.4*10-13*6.0*106 = -8254 nektons The speed of the particles are computed by the formula V=ED. This is equal to 8.5*10-6*0.15. This is equal to 84.1 Answer to question 4 Voltage = current *resistance. This implies that in this case while V is 5.0 and resistance is 1.0*103, current will be equal to 5/1.0*103, = 500 amps B the direction of the conventional current provides the electric charge movement from the positive side of the battery to its negative side as in indicated in the diagram below Answer to question 5 • This section focuses on the equivalent resistance of a circuit. The equivalent resistance will be equal to (5.0*102+1.00*103)2.
4. Calculate the density of air inside the wind tunnel and uncertainty [kg/m3]. Part 2: Calibration of a Pressure Transducer and Scannivalve 1. Zero out the micromanometer and obtain its bias. Apply this bias to all micromanometer measurements.
Physics 1408 Section E1 Standing Waves in a Vibrating Wire Callie K Partner: Miguel E Date Performed: March 20, 2012 TA: Raziyeh Y Abstract This lab had two purposes. The first was to determine the relationship between the length of a stretched wire and the frequencies at which resonance occurs. The second was to study the relationship between the frequency of vibration and the tension and linear mass density of the wire. In the first part we found the resonance, frequency and wavelength of a wire and used this data to calculate the speed of the traveling waves. For first harmonic, our wavelength was 1.200 m, found by the formula λ=2L/n.
0.0625mol/0.125M=0.5L=500mL Calculation for preparing the EDTA solution Exercise 6 a. 1L*0.02M=0.02mol 0.02mol*372.24g/mol=7.4g EDTA b. Exact molarity of 7.4448g /1.00L would be .0200 M Exercise 7 a. 0.5M*100*10-3L=0.05mol acetic acid b. 0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid.
The formula given for this problem is C = 4d -1/3b where d is the displacement in pounds and b is the beam width in feet. Also the exponent of -1/3 means the cube root of d which will be taken and the reciprocal of that number will be used in multiplication. a) First problem a 4100 Tartan sailboat capsize screening valve will be calculated. The sailboat of the beam 13.5 and a displacement of 23,245 pounds. The inches value shall be converted into decimal division.
Final exam chem 2014 Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. A central concept of the kinetic theory, one of the “big ideas” of chemistry, is the belief that a.|for every reaction there is an equal and opposite reaction.| b.|chemical reactions involve processes in which reactants produce products.| c.|the particles in matter are in a state of constant motion.| d.|every chemical process uses or produces energy, often in the form of heat.| ____ 2. In a chemical reaction, the type of products obtained is largely determined by which part of the reacting chemicals? a.|protons| b.|electrons| c.|neutrons| d.|nuclei| ____ 3.
B, C and D Electrical Circuit – an electrical device that provides a path for electrical current to flow Frequency – The cycles per second of alternating current, measured in Hertz Amplitude – the maximum difference of an alternating electric current or potential from the average value Phrase – is a position of a point in time on a waveform cycle Bit time – the time it takes for one bit to be ejected from a network interface card at a standard speed Encoding Scheme – a system used by computers to represent data as characters Unshielded twisted-pair – The word unshielded refers to the fact that UTP cables have no added shielding materials to prevent EMI problems Core- The light source on one end of the cable shines light into the core Cladding- surrounds the core, for the entire length of the cable, and reflects the light into the core Radio waves- work well for networking because as a waveform, radio waves can be changed (modulated) over time to send data. Wireless WAN- used often in mobile phone network/mobile network Wireless LAN- don’t not use cables and do use radio waves to send data Access Point – all user devices communicate only through the AP, most WLAN’s use this wireless
Using this equation I can determine the cooling time constant for the block of steel. Observations: Below you will find the data obtained from the experiment as well as the excess temperature I calculated. Time in min Actual Temperature of Block of Steel ° C Excess Temperature in ° C 0 153 128 1 133.4 108.4 2 116.7 91.7 3 102.6 77.6 4 90.7 65.7 5 80.6 55.6 6 72.1 47.1 7 64.9 39.9 8 58.7 33.7 9 53.6 28.6 10 49.2 24.2 11 45.5 20.5 12 42.3 17.3 13 39.7 14.7 14 37.4 12.4 15 35.5 10.5 16 33.9 8.9 17 32.5 7.5 18 31.4 6.4 19 30.4 5.4 20 29.6 4.6 Chart
Calculate the impedance by the measured values of voltage and current from the oscilloscope. Also calculate the impedance from = − C. Answer: Impedance, Z = 5∠0 34.65 60.48°) = 173.25 - 60.48⁰ = 85.37 Ω - j150.76 Ω Again, XC = 1/ = 1/ (6280 x 10-6) = 159.24
PHYS 222 Worksheet 5 – Electric Potential Supplemental Instruction Iowa State University Useful Equations Leader: Course: Instructor: Date: Alek Jerauld PHYS 222 Dr. Paula Herrera-Siklódy 1/24/12 Wab (Ub U a ) qq U k 0 r q U kq0 i i ri U q V k q0 r q V k i i ri dq V k r Work done by a conservative force Electric potential energy by two point charges q and q0 Potential energy of charge q0 due to the collection of charges qi Potential due to a point charge. Units: [ 1 V = 1 volt = 1 J/C] Potential due to a collection of point charges Potential due to a continuous distribution of charge Potential difference as an integral of E E in terms of V. E is the gradient of V. Vb Va E dl a b E V Diagrams