#$&*
course Phy 122
5/3 4:17Here is a practice test - I'd like to take you up on your offer to provide feedback.
Can I also say ""DANG - the modern physics section is 2 times harder and 3 times more involved than anything I've encountered so far!""
I tried to do this all from memory, but I had to refer back several times. This is gonna be a tough nut to crack on test day!" "Time and Date Stamps (logged): 14:24:22 05-03-2013 °³Ÿ±³Ÿ±±¯´Ÿ¯²Ÿ±¯°²
Principles of Physics (Phy 122) Test_4
Completely document your work and your reasoning.
You will be graded on your documentation, your reasoning, and the correctness of your conclusions.
** Write clearly in dark pencil or ink, on one side of the paper only. **
10-03-2001 20:45:55
Test should be printed using Internet Explorer. If printed from different browser check to be sure test items have not been cut off. If items are cut off then print in Landscape Mode (choose File, Print, click on Properties and check the box next to Landscape, etc.).
Signed by Attendant, with Current Date and Time: ______________________
If picture ID has been matched with student and name as given above, Attendant please sign here: _________
Instructions:
Test is to be taken without reference to text or outside notes.
Graphing Calculator is allowed, as is blank paper or testing center paper.
No time limit but test is to be taken in one sitting.
Please place completed test in Dave Smith's folder, OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va., 24212-0828 OR email copy of document to dsmith@vhcc.edu, OR fax to 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test..
Directions for Student:
Completely document your work.
Numerical answers should be correct to 3 significant figures. You may round off given numerical information to a precision consistent with this standard.
Undocumented and unjustified answers may be counted wrong, and in the case of two-choice or limited-choice answers (e.g., true-false or yes-no) will be counted wrong. Undocumented and unjustified answers, if wrong, never get partial credit. So show your work and explain your reasoning.
Due to a scanner malfunction and other errors some test items may be hard to read, incomplete or even illegible. If this is judged by the instructor to be the case you will not be penalized for these items, but if you complete them and if they help your grade they will be counted. Therefore it is to your advantage to attempt to complete them, if necessary sensibly filling in any questionable parts.
Please write on one side of paper only, and staple test pages together.
Test Problems:
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Constants:
k = 9*10^9 N m^2 / C^2 qE = 1.6 * 10^-19 C h = 6.63 * 10^-34 J s
energy of n=1 orbital in hydrogen atom: -13.6 eV k ' = 9 * 10^-7 T m / amp atomic mass unit: 1.66 * 10^-27 kg
electron mass: 9.11 * 10^-31 kg speed of light: 3 * 10^8 m/s Avogadro's Number: 6.023 * 10^-23 particles/mole
Gas Constant: R = 8.31 J / (mole K) proton mass: 1.6726 * 10^-27 kg neutron mass: 1.6749 * 10^-27 kg
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Problem Number 1
What is the power flux of light whose intensity is 4.7 watts/m ^ 2 through a square window with side 1.9 meters, if the sunlight makes an angle of 74 degrees with a 'normal' line perpendicular to the plane of the window?
.(1.9m )^2 = 3.61m^2
4.7 watts/m^2 * 3.61m^2 = 16.97watts
16.97watts * cos(74) = 4.67 watts
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Problem Number 2
A particle carrying charge 8 `microCoulombs moves at constant velocity through a combined electric and magnetic field. Both fields are perpendicular to one another and to the direction of motion of the charge, and are oriented so that the forces exerted by the two fields are in opposite directions. The electric field has strength E = 52000 N / C and the magnetic field has strength B = .02 Tesla.
Show that the charge passes through the field without a change in direction provided its velocity is v = E / B.
Show that the same is the case if the charge is 15 `microCoulombs, and explain why the change in charge had no effect on the result.
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f=qE
Force = 8* 10^-6 C * 52000N/C = .416 N
When v= E/B = 52000N/C / .02 Tesla = 2600000 m/s
F = q v B
F = 8 * 10^-6 C * 2600000 m/s * .02 Tesla = .416 N
These forces are equal in magnitude but given as in the opposite directions. They cancel out and therefore do not deflect the particle.
Since velocity is dependent on E/B these forces will always cancel each other out no matter there magnitude
Example:
f=qE
f = 15 * 10^-6 C * 52000N/C = .78 Newtons
v=E/B is STILL 52000N/C / .02 Tesla = 2600000 m/s
F = qvB
F = 15 * 10^-6 C * 2600000 m/s * .02 Tesla = .78 Newtons
again the forces are of the same magnitude but in opposite directions thus cancelling each other out.
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Problem Number 3
A magnetic field of magnitude 3.9 Tesla passes through a square loop with side 7.9 meters, with the field perpendicular to the plane of the loop. The loop is suddenly turned 90 degrees, so that its plane is parallel to the loop.
If the loop is turned in .002 seconds, what is the magnitude of the average rate of change of flux in the loop?
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When perpendicular,
flux = 3.9 Tesla * (7.9m)^2 = 243 Tm^2
When parallel, the flux is 0
So there is a change of 243 Tm^2 in .002 seconds
243 Tm^2 / .002 seconds = 121,500 Tm^2/s or 121,500 volts
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Problem Number 4
The table below depicts properties of the isotopes of the eight lightest elements. For the element of column 10, is it possible from an energy standpoint for the isotope in the fourth line to emit a neutron and change to the isotope in the third line?
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Table Does not come through in NotePad, but it basically is asking, can a Nitrogren Isotope of 15.00011 amu emit a neutron and become a nitrogen istope of 14.00307 amu?
Products would be a nucleous of 14.00307 amu +a neutron at 1.008665 amu = 15.011735 amu.
Since energy is released in the process we would expect the mass to increase.
The products have a larger mass than the original isotope. This means it IS possible from an energy standpoint.
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You can't end up with more mass-energy than you started with. If you end up with lower mass than you started with, the 'lost' mass can be there in the form of energy. But unless energy is supplied from somewhere (which can sometimes happen, but that would have to involve another particle), you can't end up with more mass.
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Problem Number 5
How many electron volts of energy are associated with a single atomic mass unit, or amu (an amu is approximately 1.66 * 10^-27 kg)? How many electron volts are associated with the mass of an electron (electron mass approximately 9.11 * 10^-31 kg)? What would be the wavelength of a single photon with this energy? Compare each wavelength with the diameter of a typical atom, which is 10^-10 m or 1 Angstrom, and with the diameter of a typical proton, which is about 10^-15 m or 1/100,000 Angstrom; this distance is also called a Fermion.
.For the amu
E= mc^2
E= 1.66 * 10^-27 kg * (3* 10^8m/s) ^2 = 1.44* 10^-10 J
1.44* 10^-10 J * (1eV/1.6*10^-19 J) = 9.00 *10^8 eV
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1.6 * 9 = 1.44, so 1.66 * 9 is a little greater.
If you correct the arithmetic you'll get 9.31 * 10^8 eV.
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For the electron
E= mc^2
E= 9.11 * 10^-31 kg * (3* 10^8m/s) ^2 = 8.199* 10^-14 J
8.199* 10^-14 J * (1eV/1.6*10^-19 J) = 5.12 *10^5 eV
Wavelenth - amu
E = h (c/lambda)
lambda = h c/E = 6.23* 10^-34 * 3.0*10^8m/s /1.44* 10^-10 J
lambda = 1.30* 10^-15 m which is about one hundred thousandth the diameter of a typical atom and on the same magnitude the diamter of a typical proton.
Wavelenth - electron energy
E = h (c/lambda)
lambda = h c/E = 6.23* 10^-34 * 3.0*10^8m/s /8.199* 10^-14 J
lambda = 2.29* 10^-12 m which is about a hundredth the diameter of a typical atom and 1000 times the size of a typical proton.
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Problem Number 6
By accurately measuring the index of refraction of light from a laser we determine that the wavelength of the light is 510 nm. How much energy is carried by each photon of this light? This light is shined on a photoelectric metal with work function .87 electron volts. A metal grid is placed near the metal and held at a variable negative potential with respect to the grid. What is the minimum magnitude of this potential which will prevent any ejected electrons from reaching the grid?
f= c/lamda
f = (3.0* 10^8m/s )/510* 10^-9m
f = 5.88 * 10^14 Hz
E = h *f
E = 6.63 * 10^ -34 Js * 5.88 * 10^14 Hz
E = 3.9 * 10^-19 J
3.9 * 10^-19 J * (1eV/ 1.6 * 10^-19J) = 2.44 eV
2.44 eV - .87 eV = 1.57 eV
Minimum magintude is 1.57 eV
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Very good but be sure to check my notes, especially on #4.
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