PHYS 222 Covering topics in Unit I EXAM #1 Fall Semester 2017 ABC Instructions:
ID: 1876354 • Letter: P
Question
PHYS 222 Covering topics in Unit I EXAM #1 Fall Semester 2017 ABC Instructions: Circle the letter corresponding to the correct answer choices. If you are sure that your answer is correct and it does not match any of those given, then write your answer on the blank space in f write in answers, correct units (SI system) must be included. Leaving off the units will cost you 1/3 rd the points for that answer, if the number is correct. Your answers should be rounded to 2 decimal places. For problems where you need to [6] #1 Point masses mi-322 kg and m2-98.6 kg are located 0.15 mm apart what is the magnitude of the force that mi exerts on mo? Answer: a. 58.3 N. b. 67.4 N. c. 72.8 N. d. 86.4% e, 94,1 N. f. [5] #2. The force between two point masses located 500 cm apart is 100 N. If the distance between the two masses is doubled the Dew force between them will be Answer: a. 25.0 N. b. 360N 40N. d. 47.5N. e, 55.6 N. [5] #3. For the masses in problem #2, if the distance remains unchanged at 500 cm but the mass of one is doubled and the other one is unchanged, the new force between the masses will be Answer a 50N b. 120N c. 200 N d. 295N e. 350 N [6] #4. Two point charges Q1--58.6 mC and Q:-15.8 mrCare separated by a distance of 75.9 min and Qi is to the left ofQa Determine the force on Q: due to Qi Answer: a. 1.06 x 10P N, toward Q. b. 2.89 x 103 N, towardQ c.6.54 x 0N, toward Q. d. 144 x 10P N, away from Q e. 3.56 x 10 N, away from Qf [S #5 At a specific location the electric field due to the two charges in problem 4 1s exactly zero where is that location relative to Qi? Answer: a. 50.0 mm, right of Q. b. 37.6 mm, left ofQ c. 25.8 mm, left of Q d. 42.1 cm, right of Q e. 58.2 mm, right of Qi Circle the FALSE answer choice, or, choices below a. If an isolated electric charge is found it must be an integral (for example, +/-1, +/-2,-3-...) multiple of 1.602 x 1019 C. b. The fundamental electric charge is e-1.602 x 10-19 C. c. Two point electric charges. Qt = +19SpC and Q,--198 pC, will attract each other d. Equipotential lines are electrie field lines that have the same electric field value e. It is possible to find exactly ¾ of an electron, if the light shines at just the right angle on a material. [5] #6. [9]s7. The electric potential in a region of space is given by V-(-3. 10 Vin|& + (6.20 vm yl-u.80 vrnz, Determine the electrie field in the fom E-EitEj+Ek at the point (0.0 m,+5.0 m 3.0m). b. E = (30.0 Vinit-(140 V1ny + (3 Vinik. d. E = (4.50 Virn)I + (47.2 Viny-(3 Vink c. e. E = (3.10 V.,mi-162.0 Viny (48.6 vinik. E-(-17.0 Vini . (28.1 Vimj-(24.0 Vinik [6] #8. Determine the net electric flux through the surface shown by the broken line. -96 -75e -60e +86e. Answer: -94e +58e [6] #9. At a certain location the electric field is given by E-(+51-71+ 9 k) Vm. Determine the electric flux through a surface whose area vector at that location is A-(-15.0j 36.0 k) m Answer: a. 66.5 V.m b. 124Vm c. 174 V.m d.-219 Vm e. -257 V.m f [6] #10. what is the force (magnitude and direction) on an electron that is placed at a point where the magnitude of an electric field is E- 15.8 (V/m)? Answer Note: For the Direction, circle the correct answer in the parentbeses PHYS 222 Direction: Same, or, opposite) to E EXAM #1 Spring Semester 2014 ABC Name:Explanation / Answer
6) The magnitude of the force exerted is given by:
F = Gm1m2/r2
= (6.67*10-11)(322*98.6)/(0.15*10-3m)2
= 94.1 N (option e)
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