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06homework05

06homework05 - Husain Zeena Homework 5 Due 4:00 am Inst...

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Husain, Zeena – Homework 5 – Due: Feb 23 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 38 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points A strong lightning bolt transfers an electric charge of about 11 C to Earth (or vice versa ). How many electrons are transferred? Correct answer: 6 . 86566 × 10 19 . Explanation: Given : q = 11 C and q e = - 1 . 60218 × 10 - 19 C . The charge is proportional to the number of electrons, so q = n q e n = q q e = - 11 C - 1 . 60218 × 10 - 19 C = 6 . 86566 × 10 19 . The charge on the electron is negative. The direction of current flow in lightning can be either from Earth to the clouds or vice versa . 002 (part 2 of 2) 10 points Assume: One mole of water has a mass of 18 . 1 g / mol. If each water molecule donates one electron, how much water is ionized in the lightning? Correct answer: 0 . 00206353 g. Explanation: Given : M = 18 . 1 g / mol and N a = 6 . 02214 × 10 23 / mol Mass is proportional to the number of atoms in a substance, so for m grams in N atoms in the water and M grams in N a atoms in one mole, we have m M = N N a N = m M N a Hence the total mass M total of the water ion- ized by the lightning bolt is give by M total = n H 2 O m H 2 O = n e M H 2 O N A = ( 6 . 86566 × 10 19 ) (18 . 1 g / mol) (6 . 02214 × 10 23 / mol) = 0 . 00206353 g , Since every water molecule donates one elec- tron to the discharge, the total number of transferred electrons n e equals the total num- ber n H 2 O of ionized water molecules, so n e = n H 2 O . 003 (part 1 of 2) 10 points The quantity of charge passing through a sur- face of area 2 . 89 cm 2 varies with time as q = q 1 t 3 + q 2 t + q 3 , where q 1 = 4 . 9 C / s 3 , q 2 = 6 . 1 C / s, q 3 = 7 . 4 C, and t is in seconds. What is the instantaneous current through the surface at t = 0 . 9 s? Correct answer: 18 . 007 A. Explanation: Given : q 1 = 4 . 9 C / s 3 , q 2 = 6 . 1 C / s , q 3 = 7 . 4 C , and t = 0 . 9 s . I d q dt = 3 q 1 t 2 + q 2 = 3 ( 4 . 9 C / s 3 ) (0 . 9 s) 2 + 6 . 1 C / s = 18 . 007 A . 004 (part 2 of 2) 10 points What is the value of the current density at

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Husain, Zeena – Homework 5 – Due: Feb 23 2004, 4:00 am – Inst: Sonia Paban 2 t = 0 . 9 s? Correct answer: 62308 A / m 2 . Explanation: Given : a = 2 . 89 cm 2 = 0 . 000289 m 2 and t = 0 . 9 s . J I A = 18 . 007 A 0 . 000289 m 2 = 62308 A / m 2 . 005 (part 1 of 5) 10 points The mass density of Cu is 8 . 92 g / cm 3 , its molar mass is 63 . 5 g / mol, and Avogadro’s number is 6 . 02214 × 10 23 atoms / mole. The electron mass is 9 . 10939 × 10 - 31 kg, and the resistivity of copper is 1 . 7 × 10 - 8 Ω · m. A current of 25 A exists in a copper (Cu) wire which has a diameter of 5 mm. Assume: Each atom of copper has one conduction band, and the average thermal speed of an electron (in the free electron model of conductivity v = r k T m ) = 1 × 10 6 m / s . What is the current density, J ? Correct answer: 1 . 27324 × 10 6 A / m 2 . Explanation: Given : I = 25 A and r = 2 . 5 mm = 0 . 0025 m . The current density is given by J I A = I π r 2 .
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06homework05 - Husain Zeena Homework 5 Due 4:00 am Inst...

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