Exam's Compiled - Kapoor (mk9499) oldmidterm 01 Turner...

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Kapoor (mk9499) – oldmidterm 01 – Turner – (60230) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Two spheres, Fastened to “pucks”, are rid- ing on a Frictionless airtrack. Sphere “1” is charged with 1 nC, and sphere “2” is charged with 3 nC. Both objects have the same mass. 1 nC is equal to 1 × 10 - 9 C. As they repel, 1. sphere “2” accelerates 3 times as Fast as sphere “1”. 2. sphere “1” accelerates 9 times as Fast as sphere “2”. 3. sphere “2” accelerates 9 times as Fast as sphere “1”. 4. they have the same magnitude oF acceler- ation. correct 5. they do not accelerate at all, but rather separate at constant velocity. 6. sphere “1” accelerates 3 times as Fast as sphere “2”. Explanation: The Force oF repulsion exerted on each mass is determined by F = 1 4 π± 0 Q 1 Q 2 r 2 = ma where r is the distance between the centers oF the two spheres. ± - F 12 ± = ± - F 21 ± Since both spheres have the same mass and are subject to the same Force, they have the same acceleration. 002 10.0 points Three equal charges oF 3 μ C are in the x - y plane. One is placed at the origin, another is placed at (0.0, 99 cm), and the last is placed at (85 cm, 0.0). The Coulomb constant is 9 × 10 9 Nm 2 / C 2 . Calculate the magnitude oF the Force on the charge at the origin. Correct answer: 0 . 13928 N. Explanation: Let : q =3 μ C = 3 × 10 - 6 C , ( x 1 ,y 1 ) = (0 , 99 cm) = (0 , 0 . 99 m) , ( x 2 2 ) = (85 cm , 0) = (0 . 85 m , 0) and ( x 0 0 ) = (0 , 0) . The electric felds are E y = k e q x 2 1 + y 2 1 = ( 9 × 10 9 2 / C 2 )( 3 × 10 - 6 C ) 0 + (0 . 99 m) 2 = 27548 . 2N / C , and E x = k e q x 2 2 + y 2 2 = ( 9 × 10 9 2 / C 2 3 × 10 - 6 C ) (0 . 85 m) 2 +0 = 37370 . / C , Thus E = ± E 2 x + E 2 y and F = qE = q ± E 2 x + E 2 y = ( 3 × 10 - 6 C ) · ± (27548 . / C) 2 + (37370 . / C) 2 = 0 . 13928 N . 003 10.0 points Two identical small charged spheres hang in equilibrium with equal masses as shown in the fgure. The length oF the strings are equal and the angle (shown in the fgure) with the vertical is identical. The acceleration oF gravity is 9 . 8m / s 2 and the value oF Coulomb’s constant is 8 . 98755 × 10 9 2 / C 2 .
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Kapoor (mk9499) – oldmidterm 01 – Turner – (60230) 2 0 . 15 m 5 0 . 03 kg 0 . 03 kg Find the magnitude of the charge on each sphere. Correct answer: 4 . 4233 × 10 - 8 C. Explanation: Let : L =0 . 15 m , m . 03 kg , and θ =5 . L a θ m m q q From the right triangle in the ±gure above, we see that sin θ = a L . Therefore a = L sin θ = (0 . 15 m) sin(5 ) . 0130734 m . The separation of the spheres is r =2 a = 0 . 0261467 m . The forces acting on one of the spheres are shown in the ±gure below. θ θ m g F T e T sin θ T cos θ Because the sphere is in equilibrium, the resultant of the forces in the horizontal and vertical directions must separately add up to zero: ± F x = T sin θ - F e ± F y = T cos θ - mg . From the second equation in the system above, we see that T = cos θ , so T can be eliminated from the ±rst equation if we make this substitution. This gives a value F e = tan θ = (0 . 03 kg) ( 9 .
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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Exam's Compiled - Kapoor (mk9499) oldmidterm 01 Turner...

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