1 - karna (pk4534) – HW01 – li – (59050) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: karna (pk4534) – HW01 – li – (59050) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The five thousand billion freely moving elec- trons in a penny repel one another. Why don’t they fly off the penny? 1. The electrons attract each other. 2. They cause a jam when they try to fly away. 3. They are attracted to the five thou- sand billion positively charged protons in the atomic nuclei of atoms in the penny. correct 4. The shell of the penny prevents the elec- trons from flying. 5. They don’t have enough speed. Explanation: The electrons are bound to the nuclei. 002 10.0 points How can a charged atom (an ion) attract a neutral atom? 1. The charged atom can produce secondary electrons to interact with the neutral atom and make it positively charged or negatively charged. 2. An ion polarizes a nearby neutral atom, so that the part of the atom nearer to the ion acquires a charge opposite to the charge of the ion, and the part of the atom farther from the ion acquires a charge of the same sign as the ion. correct 3. The charged atom can emit x rays to induce ionization of the neutral atom. 4. The charged atom can hit the neutral atom and make it positively charged or nega- tively charged. Explanation: An ion polarizes a nearby neutral atom, so that the part of the atom nearer to the ion acquires a charge opposite to the charge of the ion, and the part of the atom farther from the ion acquires a charge of the same sign as the ion. 003 10.0 points A particle of mass 41 g and charge 62 μ C is released from rest when it is 18 cm from a second particle of charge − 20 μ C. Determine the magnitude of the initial ac- celeration of the 41 g particle. Correct answer: 8389 . 42 m / s 2 . Explanation: Let : m = 41 g , q = 62 μ C = 6 . 2 × 10 − 5 C , d = 18 cm = 0 . 18 m , Q = − 20 μ C = − 2 × 10 − 5 C , and k e = 8 . 9875 × 10 9 . The force exerted on the particle is F = k e | q 1 | | q 2 | r 2 = ma bardbl vectora bardbl = k e bardbl vectorq bardblbardbl vector Q bardbl md 2 = k e vextendsingle vextendsingle 6 . 2 × 10 − 5 C vextendsingle vextendsingle vextendsingle vextendsingle − 2 × 10 − 5 C vextendsingle vextendsingle (0 . 041 kg) (0 . 18 m 2 ) = 8389 . 42 m / s 2 . 004 10.0 points Two spheres, fastened to “pucks”, are rid- ing on a frictionless airtrack. Sphere “1” is charged with 2 nC, and sphere “2” is charged with 6 nC. Both objects have the same mass. 1 nC is equal to 1 × 10 − 9 C. As they repel, 1. they have the same magnitude of acceler- ation. correct 2. they do not accelerate at all, but rather separate at constant velocity. karna (pk4534) – HW01 – li – (59050) 2 3. sphere “2” accelerates 3 times as fast as sphere “1”....
View Full Document

This note was uploaded on 11/03/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 10

1 - karna (pk4534) – HW01 – li – (59050) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online