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
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.
Amounts Variable overhead: Price variance $ (0%) (0%) Efficiency variance $ (0%) (0%) Fixed overhead: Price variance $ (0%) (0%) ________________________________________ P16-45 Overhead Variances (L.O. 5, 6) Lima Parts, Inc., shows the following overhead information for the current period: Actual overhead incurred $ 29,400 2/3 of which is variable Budgeted fixed overhead $ 8,640 per
Trial 1 V (v) | I (A) | 0.5 | .02 | 1.0 | .04 | 1.5 | .06 | 2.0 | .08 | 2.5 | .10 | 3.0 | .12 | 3.5 | .14 | 4.0 | .16 | 4.5 | .18 | 5.0 | .20 | Table 1. This table shows the magnitude found when using the Power Resistor portion of the experiment. Trial 2 V (v) | I (A) | 2.0 | .05 | 4.0 | .08 | 6.0 | .08 | 8.0 | .10 | 10.0 | .11 | 12.0 | .12 | 14.0 | .12 | 16.0 | .13 | 18.0 | .14 | 20.0 | .14 | Table 2. This table shows the magnitude found when using the Light Bulb portion of the experiment Trial 3 V (v) | I (A) | 2.0 | 0.1 | 4.0 | 0.19 | 6.0 | 0.29 | 8.0 | 0.38 | 10.0 | 0.50 | 12.0 | 0.60 | 14.0 | 0.70 | 16.0 | 0.80 | 18.0 | 0.90 | 20.0 | 1.00 | Table 3. This table shows the magnitude found when using the rheostat portion of the experiment.
The SD for N=100: | 0.345 | 8a. Perentile score for 99°F: | 0.89 | 8b. z score for 99° F: | 1.29 | 8c. Is a body temperature of 99.00°F “unusual”? Why or why not?
TOPIC 8 Chi-Square goodness-of-fit test Problem 12.1 Use a chi-square goodness-of-fit to determine whether the observed frequencies are distributed the same as the expected frequencies (α = .05) Category | fo | fe | 1 | 53 | 68 | 2 | 37 | 42 | 3 | 32 | 33 | 4 | 28 | 22 | 5 | 18 | 10 | 6 | 15 | 8 | Step 1 Ho: The observed frequencies are distributed the same as the expected frequencies Ha: The observed frequencies are not distributed the same as the expected frequencies Step 2 df = k – m – 1 Step 3 α = 0.05 x 2 0.05, 5df = 11.0705 Step 4 Reject Ho if x 2 > 11.0705 Category | fo | fe | | 1 | 53 | 68 | | 2 | 37 | 42 | | 3 | 32 | 33 | | 4 | 28 | 22 | | 5 | 18 | 10 | | 6 | 15 | 8
A second point charge q2 = -4.30 µC moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m, y = 0.250 m. How much work is done by the electric force on q2? W (U b U a ) Ua Ub kq1q2 9(109 )(2.4)(106 )( 4.3)(106 ) 0.6192 J x1 0.150 kq1q2 x2 2 y2 2 9(109 )(2.4)(106 )( 4.3)(106 ) 0.252 0.252 0.2627 J W (U b U a ) ((0.2627) (0.6192)) 0.3565 J 2) Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge? KEi U i KE f U f KE f U i U
a. Patient with primary hypothyroidism Patient 2 t4 is 0.01 and its concentration is low. Laboratory Report/ Eriquita Hampton/ Homeostatic Imbalances of Thyroid Function/ Bela Struckman/ 12.04.2014/ Page [1] of [3] b.
Experiment Laboratory techniques and MeasureMents table 2: Prefixes used in the SI System. Prefix penta tera giga mega kilo hecto deka ----deci centi milli micro nano pico femto Symbol P t G m k h da ---d c m m n p f Meaning 1,000,000,000,000,000 1,000,000,000,000 1,000,000,000 1,000,000 1,000 100 10 1 0.1 0.01 0.001 0.000001 0.000000001 0.000000000001 0.000000000000001 Exponential Notation 1015 1012 109 106 103 102 101 100 10-1 10-2 10-3 10-6 10-9 10-12 10-15 We will now discuss each fundamental SI system unit of measure and the laboratory techniques and equipment that scientists use to obtain each measurement. length: length is defined as the distance (amount of space) of an object from end to end. The SI system unit of length is the meter (m) and was originally intended to represent one ten-millionth of the distance between the North Pole and the equator. However, over time the definition of the meter changed.
| Variable cost/ unit | Total Fixed cost | Total cost | Total cost/unit | Direct Materials | =450-300 600-400=0.75 | - | =375 | =0.75 | Direct Labour | =750-500 600-400=1.25 | - | =625 | =1.25 | Indirect Labour | = 220-180 600-400=0.2 | =180- (0.2×400)=100 | =200 | =0.4 | Indirect Materials | - | =300 | =300 | =0.6 | Electriciy | =135-115 600-400=0.1 | =115-(0.1×400)=75 | =125 | =0.25 | Factory Insurance | - | =125 | =125 | =0.25 | Other Overhead | =410-310 600-400=0.5 | =310-(0.5×400)=110 | =360 | =0.72 | Total | =2.8 | =710 | =2110 | =4.22 | b. Work out the detail on how Lee High put together his ‘useful data on Great Heath’ and the standardized cost information based on 500 units per week. Manufacturing cost 4.22 Administration cost Variable cost (commission) (7×10%) 0.7 Fixed cost (781/500) 1.56 2.26 6.48 Funny error 0.12 Total cost/ unit 6.60 The cost of goods sold per unit tended to fall when the sales increased because of the characteristic of the fixed costs. Fixed costs remained constant over wide ranges of activity for a specified time period. They were not affected by changes in activity.