*ratings for typed answers ONLY*
Answer the following question: How would your results differ if you performed the experiment on Mars? with the given information below!
Using the slope of your T2 versus L plot, determine the acceleration due to gravity.
The slope of the graph = 0.042924
Acceleration due to gravity 39.47/0.042924=939,761
Acceleration due to gravity = 9.39 m/s2
Based on your results, please answer the following questions:
Based on our calculus, our result equals 9.39 m/s2 is close to the value of 9.81 m/s2. The explanation for the small difference is represented by potential errors that took place during the experiment. In most cases, this is due to human error. This can include parallax error that took place when the measurement of lengths was done or a mistake in recording the period of time.
It is important to understand that the period T is dependent on the length of the string. As we can see, the higher the length is, the longer the value of T is. Because of that, the period T won't be dependent on the mass of the initial angular displacement.
An important thing during this experiment is to keep the string as straight as possible during the entire period of motion. In order to do that, it needs to use a relatively high mass to straighten the line. Using only one washer can lead to an incorrect result. This is because one washer represents a small weight, and the air resistance will be more effective.
In: Physics
Dataset #2 – Star War Film Data
Description: Weekly domestic box office revenues for the 8 Star War films
Research ‘Question’: Find a ‘best’ linear model to predict Star War revenue/day using the number of theaters, number of weeks since release, film number, and release year.
| theaters | weeknum | film | year | revperday |
| 3672 | 1 | IV | 1977 | 18498679.7 |
| 3672 | 2 | IV | 1977 | 9505314.86 |
| 3672 | 3 | IV | 1977 | 4127697.71 |
| 3672 | 4 | IV | 1977 | 2632591 |
| 3422 | 5 | IV | 1977 | 1950438.14 |
| 3311 | 6 | IV | 1977 | 2521766.29 |
| 3186 | 7 | IV | 1977 | 2831227.86 |
| 2681 | 8 | IV | 1977 | 1023363.71 |
| 2170 | 9 | IV | 1977 | 652710.714 |
| 1851 | 10 | IV | 1977 | 566439 |
| 1202 | 11 | IV | 1977 | 250623.714 |
| 907 | 12 | IV | 1977 | 179533.714 |
| 505 | 13 | IV | 1977 | 102494.857 |
| 311 | 14 | IV | 1977 | 74403.1429 |
| 206 | 15 | IV | 1977 | 44651.5714 |
| 215 | 16 | IV | 1977 | 46953.5714 |
| 228 | 17 | IV | 1977 | 54924.2857 |
| 172 | 18 | IV | 1977 | 29591.1429 |
| 291 | 19 | IV | 1977 | 76476.1429 |
| 270 | 20 | IV | 1977 | 59581 |
| 160 | 21 | IV | 1977 | 41030.1429 |
| 111 | 22 | IV | 1977 | 28579.4286 |
| 57 | 23 | IV | 1977 | 22707.5714 |
| 43 | 24 | IV | 1977 | 17242.4286 |
| 40 | 25 | IV | 1977 | 11668.7143 |
| 30 | 26 | IV | 1977 | 9229 |
| 3682 | 1 | V | 1980 | 15161652.6 |
| 3682 | 2 | V | 1980 | 8844278.29 |
| 3682 | 3 | V | 1980 | 5120454.57 |
| 3387 | 4 | V | 1980 | 1772898.57 |
| 3025 | 5 | V | 1980 | 1165040.57 |
| 2505 | 6 | V | 1980 | 1340427.71 |
| 2505 | 7 | V | 1980 | 1944470 |
| 2015 | 8 | V | 1980 | 799467 |
| 1550 | 9 | V | 1980 | 421755.857 |
| 1077 | 10 | V | 1980 | 303789.143 |
| 783 | 11 | V | 1980 | 142854.857 |
| 502 | 12 | V | 1980 | 85785.1429 |
| 352 | 13 | V | 1980 | 52545.1429 |
| 441 | 14 | V | 1980 | 70452.4286 |
| 388 | 15 | V | 1980 | 45788.2857 |
| 388 | 16 | V | 1980 | 41332.7143 |
| 360 | 17 | V | 1980 | 39414.5714 |
| 205 | 18 | V | 1980 | 24388.8571 |
| 151 | 19 | V | 1980 | 17734.5714 |
| 95 | 20 | V | 1980 | 14462.7143 |
| 80 | 21 | V | 1980 | 12256.4286 |
| 72 | 22 | V | 1980 | 4412 |
| 15 | 23 | V | 1980 | 786.285714 |
| 7 | 24 | V | 1980 | 455.285714 |
| 3855 | 1 | VI | 1983 | 17580664.1 |
| 3855 | 2 | VI | 1983 | 7119019.71 |
| 3805 | 3 | VI | 1983 | 3913192.71 |
| 3004 | 4 | VI | 1983 | 2412629 |
| 2725 | 5 | VI | 1983 | 1652119.43 |
| 2002 | 6 | VI | 1983 | 977608.429 |
| 1460 | 7 | VI | 1983 | 643752.429 |
| 1008 | 8 | VI | 1983 | 404027.429 |
| 605 | 9 | VI | 1983 | 240410.429 |
| 409 | 10 | VI | 1983 | 169831.286 |
| 310 | 11 | VI | 1983 | 107789.429 |
| 248 | 12 | VI | 1983 | 80801.4286 |
| 391 | 13 | VI | 1983 | 95609.8571 |
| 391 | 14 | VI | 1983 | 90454.4286 |
| 321 | 15 | VI | 1983 | 38485 |
| 228 | 16 | VI | 1983 | 29893 |
| 246 | 17 | VI | 1983 | 25054 |
| 164 | 18 | VI | 1983 | 11661.4286 |
| 119 | 19 | VI | 1983 | 9036 |
| 74 | 20 | VI | 1983 | 8862.57143 |
| 55 | 21 | VI | 1983 | 7250 |
| 55 | 22 | VI | 1983 | 5731.71429 |
| 3858 | 1 | I | 1999 | 20897581.3 |
| 3858 | 2 | I | 1999 | 9015073 |
| 3858 | 3 | I | 1999 | 3487897.43 |
| 3325 | 4 | I | 1999 | 1834563.57 |
| 2750 | 5 | I | 1999 | 1438515.14 |
| 2424 | 6 | I | 1999 | 1818900.29 |
| 2316 | 7 | I | 1999 | 1315771.29 |
| 1555 | 8 | I | 1999 | 510037.571 |
| 1003 | 9 | I | 1999 | 345916.714 |
| 560 | 10 | I | 1999 | 159016.429 |
| 340 | 11 | I | 1999 | 96117.5714 |
| 245 | 12 | I | 1999 | 69097 |
| 160 | 13 | I | 1999 | 49419.4286 |
| 441 | 14 | I | 1999 | 136217 |
| 422 | 15 | I | 1999 | 93123.1429 |
| 331 | 16 | I | 1999 | 57197.7143 |
| 231 | 17 | I | 1999 | 39329.1429 |
| 191 | 18 | I | 1999 | 29226.5714 |
| 140 | 19 | I | 1999 | 22458.7143 |
| 89 | 20 | I | 1999 | 14974.7143 |
| 4285 | 1 | II | 2002 | 19483946.1 |
| 4285 | 2 | II | 2002 | 7050087.71 |
| 4005 | 3 | II | 2002 | 3828435.43 |
| 3125 | 4 | II | 2002 | 2158583 |
| 2585 | 5 | II | 2002 | 1212925.71 |
| 1955 | 6 | II | 2002 | 817540.571 |
| 1322 | 7 | II | 2002 | 488799.571 |
| 1017 | 8 | II | 2002 | 417103.143 |
| 775 | 9 | II | 2002 | 193287.571 |
| 589 | 10 | II | 2002 | 143490.429 |
| 320 | 11 | II | 2002 | 59758.8571 |
| 241 | 12 | II | 2002 | 41315.4286 |
| 408 | 13 | II | 2002 | 74103.8571 |
| 377 | 14 | II | 2002 | 54086.4286 |
| 283 | 15 | II | 2002 | 38864.1429 |
| 225 | 16 | II | 2002 | 27574.1429 |
| 159 | 17 | II | 2002 | 18940 |
| 105 | 18 | II | 2002 | 14270.4286 |
| 90 | 19 | II | 2002 | 9984.85714 |
| 56 | 20 | II | 2002 | 8214.28571 |
| 52 | 21 | II | 2002 | 4788.28571 |
| 38 | 22 | II | 2002 | 2020.85714 |
| 4325 | 1 | III | 2005 | 21314847.9 |
| 4455 | 2 | III | 2005 | 6561318.43 |
| 4393 | 3 | III | 2005 | 3879632 |
| 3455 | 4 | III | 2005 | 1973952.71 |
| 2771 | 5 | III | 2005 | 1146060.29 |
| 1936 | 6 | III | 2005 | 718753.857 |
| 1508 | 7 | III | 2005 | 474352.286 |
| 1091 | 8 | III | 2005 | 403442.857 |
| 744 | 9 | III | 2005 | 173298.571 |
| 415 | 10 | III | 2005 | 78098.7143 |
| 301 | 11 | III | 2005 | 51525.8571 |
| 190 | 12 | III | 2005 | 33442.8571 |
| 505 | 13 | III | 2005 | 84180.1429 |
| 356 | 14 | III | 2005 | 51179.8571 |
| 245 | 15 | III | 2005 | 33814.8571 |
| 201 | 16 | III | 2005 | 21102 |
| 135 | 17 | III | 2005 | 17775.7143 |
| 95 | 18 | III | 2005 | 11938.8571 |
| 44 | 19 | III | 2005 | 7837.85714 |
| 44 | 20 | III | 2005 | 6345.28571 |
| 36 | 21 | III | 2005 | 3118.28571 |
| 23 | 22 | III | 2005 | 1052.42857 |
| 4125 | 1 | VII | 2015 | 24281289.7 |
| 4125 | 2 | VII | 2015 | 8218801.86 |
| 4125 | 3 | VII | 2015 | 3098252 |
| 3577 | 4 | VII | 2015 | 1644693.14 |
| 1840 | 5 | VII | 2015 | 1302432.86 |
| 1732 | 6 | VII | 2015 | 1294747 |
| 1732 | 7 | VII | 2015 | 918122.286 |
| 1507 | 8 | VII | 2015 | 442270.857 |
| 941 | 9 | VII | 2015 | 291175.571 |
| 725 | 10 | VII | 2015 | 168580.857 |
| 465 | 11 | VII | 2015 | 109324.714 |
| 365 | 12 | VII | 2015 | 71774.2857 |
| 409 | 13 | VII | 2015 | 93213.2857 |
| 321 | 14 | VII | 2015 | 77634.8571 |
| 303 | 15 | VII | 2015 | 45363.7143 |
| 208 | 16 | VII | 2015 | 30144.8571 |
| 122 | 17 | VII | 2015 | 20494.5714 |
| 94 | 18 | VII | 2015 | 14027.7143 |
| 85 | 19 | VII | 2015 | 12463.4286 |
| 66 | 20 | VII | 2015 | 8202.42857 |
| 4375 | 1 | VIII | 2017 | 32302438.4 |
| 4375 | 2 | VIII | 2017 | 10059634.3 |
| 4145 | 3 | VIII | 2017 | 4872357.86 |
| 3175 | 4 | VIII | 2017 | 2777846.71 |
| 2414 | 5 | VIII | 2017 | 1630078.29 |
| 1738 | 6 | VIII | 2017 | 963457.571 |
| 1328 | 7 | VIII | 2017 | 558613 |
| 1092 | 8 | VIII | 2017 | 564588.286 |
| 810 | 9 | VIII | 2017 | 196717.429 |
| 601 | 10 | VIII | 2017 | 136677.857 |
| 320 | 11 | VIII | 2017 | 76497 |
| 252 | 12 | VIII | 2017 | 53219.8571 |
| 407 | 13 | VIII | 2017 | 86566.5714 |
| 330 | 14 | VIII | 2017 | 57112.1429 |
| 240 | 15 | VIII | 2017 | 35131 |
| 163 | 16 | VIII | 2017 | 22387.2857 |
| 225 | 17 | VIII | 2017 | 21222.2857 |
| 85 | 18 | VIII | 2017 | 10420.1429 |
| 78 | 19 | VIII | 2017 | 5208.14286 |
In: Statistics and Probability
Question to answer after EXPERIMENT OF SPECTROPHOTOMETRIC DETERMINATION OF TRACE Fe
Principles The purpose of this experiment is to determine the concentration of iron present as a trace constituent (ppm) in an aqueous sample by a spectrophotometric method. One of the most sensitive methods for the determination of iron involves the formation of the orange-red iron (II) orthophenanthroline complex. This complex has an absorption maximum at 508 nm, is stable over a long period of time, and follows Beer's Law quite faithfully.
The chemistry involved is straightforward. Hydroxylamine hydrochloride is used to reduce all Fe3+ present to Fe2+. Since orthophenanthroline (abbreviated as phen) is a weak base and an acid medium must be used in order to prevent precipitation of the iron as Fe(OH)2, the principal complexing species is the phenanthrolium ion (phenH+ ). The complex formation reaction:
Fe2+ + 3phenH+ Fe(phen)3 2+ + 3H+
has an equilibrium constant of 2.5 x 106 , so complex formation may be considered quantitative. Although pH control is not essential, too high of an H+ concentration (pH<2) can inhibit formation of the complex, while too high of a pH can cause the loss of iron by precipitation. Thus, the solution is buffered to be to about a pH of 3.5.
Since the system obeys Beer's Law, above, a plot of absorbance at 508 nm versus iron concentration yields a straight line. Thus, one can prepare a set of standard solutions of known concentrations, establish a straight line, and prepare the unknown solution in exactly the same way. If the absorbance of the unknown solution is measured, the unknown concentration may be determined from the analytical curve generated with the standard solutions.
Questions:
1. The ferrous ammonium sulfate used as a standard in this experiment is not of primary standard quality, yet you were not instructed to standardize this reagent 81 before use. Why is it not necessary to standardize this analytical reagent before use in this experiment? Explain.
2. Describe two reasons why deviations to Beer's Law (i.e. nonlinearity) can occur.
3. The sensitivity of spectrophotometric methods can greatly exceed sensitivities attainable with classical wet chemical methods. How do the Fe concentrations analyzed in this spectrophotometric experiment compare with the Fe concentrations used in the redox titration experiment? Be quantitative!
In: Chemistry
Write a class encapsulating the concept of a corporate name (for example, IBM), assuming a corporate name has the following attribute: the corporate name. Include a constructor, the accessors and mutators, and methods toString() and equals(). Also include and method that returns a potential domain name by adding a www. at the beginning and .com at the end of the corporate name (for instance, if the corporate name is IBM, that method should return www.ibm.com). Write a client class to test all the methods in your class.
In: Computer Science
Complete Answer Must Compile in Visual Code in template Form Code, Must Reflect in UI.
1. Add a new text box - Mavenlink Project Name
2. Once we enter project name and project code , on tab change Mavenlink Project Name should be set as
Project Code - Project Name
Example :
Project Code = T1002
Project Name = TEST123
So in this case Mavenlink Project Name will be T1002 - TEST123
In: Computer Science
In an ADC, assume that Vref is connected to 2.3 V. Find the following. Assume an 9-bit option.
a-) Step Size = Answer mV (answer must have 3 digits after the decimal point, rounded).
b-) D8..D0 values if Vin = 1.39V Answer (answer should be rounded and then converted)
c-) D8..D0 values if Vin = 2.17V Answer (answer should be rounded and then converted)
d-) Vin if D8..D0 values are 001010101 => Answer V (answer must have 3 digits after the decimal point, rounded).
e-) Vin if D8..D0 values are 111011011 => Answer V (answer must have 3 digits after the decimal point, rounded).
In: Electrical Engineering
Using a light source with a power density of 500 W /
m2 for a PV module with a surface area of 0.92 m2, the measured
values at the Tcell = 55 ˚C are given as follows.
The current and voltage value corresponding to the maximum power
point is Vmpp = 30 V and Impp = 1.45.
The open circuit voltage is Voc 38 V and the short circuit
current is Isc 1.70.
The temperature coefficient for this module Voc is 0.16 V / ° C
(ignore for Isc).
According to this, calculate the following.
a) What is the efficiency of this module for the specified
measurement conditions?
b) Calculate the Voc, Isc, Vm, Im, and Pmax values for this module
in standard test measurement conditions (STC).
c) What is the efficiency of the module in STC?
In: Physics
(a) Find the equilibrium solution, or critical point, of the given system.
(b) Use a computer to draw a direction field and phase portrait centered at the critical point.
(c) Describe how solutions of the system behave in the vicinity of the critical point.
x′ =−0.25x−0.75y+8, y′ =0.5x+y−11.5
(d) Let x= xc+u and y= yc+v, where xc and yc give the critical point you found in (a). Plug these into the system and show that you obtain a homogeneous system u′ = Au for u = (u v)T .
(e) Solve the resulting homogeneous system for u and v, and show that the solutions you obtain match the phase portrait that you generated in (b).
In: Advanced Math
A ballistic pendulum consists of a large heavy mass, M on the end of a very light rod of length L. The rod is free to pivot at the top and the mass is attached at the bottom, so it hangs like a pendulum. A second mass, m, is fired horizontally at speed v straight into the large mass, and they stick together. The pendulum swings to a maximum angle ?m.
a)Show that you can measure v by measuring ?m, that is obtain an expression for v in terms of M, m, L, g and ? m.
b)Look up the mass and speed of a bullet, and figure out how mass
ive M should be if L is about 0.25 m and we want a maximum angular displacement no greater than 45?.
In: Physics
We have to calculate the value of Power of an electric DC motor IN WATTS made from scratch. The only known values that we have are: The Voltage (V) which is 9V and the current (I) that we are using which is 3 A, also the value of "n" which is n= 90.16 m (the number of turns of the wire used in the motor)
I already asked here for this problem and someone answered this:
Voltage = 9 V, current = 3 A, number of turns = n = 90.16
Power of motor = n*V*I = 90.16*9*3=2434.32 Watt
But I don’t think this is correct since the motor has power in and power out. Can someone else help? Thank you!
In: Physics