### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Completed Homework CH 25-29 PHYS 1444 - 001

UTA

GPA 4.0

### View Full Document

## 68

## 0

## Popular in GENERAL TECHNICAL PHYSICS II

## Popular in Physics 2

This 5 page Bundle was uploaded by ernest on Wednesday March 23, 2016. The Bundle belongs to PHYS 1444 - 001 at University of Texas at Arlington taught by Suman Satyal, Suresh C Sharma in Fall 2015. Since its upload, it has received 68 views. For similar materials see GENERAL TECHNICAL PHYSICS II in Physics 2 at University of Texas at Arlington.

## Similar to PHYS 1444 - 001 at UTA

## Reviews for Completed Homework CH 25-29

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 03/23/16

martinez (m6688) – Electric Current – satyal – (14442016S) 1 This print-out should have 21 questions. = a t − b t + ct . Multiple-choice questions may continue on 3 1 2 1 1 the next column or page – ﬁnd all choices before answering. 002 10.0 points The compressor on an air conditioner draws 001 10.0 points 20.0 A when it starts up. The current If the start-up time is 0.75 s, how much I = at − bt + c charge passes a cross-sectional area of the circuit in this time? in a section of a conductor depends on time. What quantity of charge moves across the Correct answer: 15 C. section of the conductor from t = 0 s to t = t1? Explanation: 1. q = 2at 1 b Let : I = 20.0 A and 3 b 2 ∆t = 0.75 s. 2. q = at 1 t1+ ct 1 2 3 3. q = at 1 bt +2ct 1 ∆Q a 3 b 2 I = 4. q = t1− t1+ c ∆t 3 2 2 ∆Q = I ∆t = (20 A)(0.75 s) = 15 C . 5. q = 2at 1 bt 1 a 3 b 2 6. q = t1− t1+ ct 1orrect 003 10.0 points 3 2 The current in a lightning bolt is 6.8 × 10 A. 2 7. q = 3at 1 − b + c How much charge passes through a cross- sectional area of the lightning bolt in 4.8 s? 8. q = at − bt + c 1 1 6 Correct answer: 3.264 × 10 C. 9. q = 3at 2 − 2bt 1 c 1 Explanation: 10. q = 2at 1 3bt + 1t 1 5 Explanation: Let : I = 6.8 × 10 A and The unit of current is Coulomb per second: ∆t = 4.8 s. I = dq or dq = I dt. dt ∆Q To ﬁnd the total charge that passes through I = the conductor, one must integrate the current ∆t over the time interval. ∆Q = I ∆t = (6.8 × 10 A)(4.8 s) = 3.264 × 10 C . Z 1 Z t1 q = dq = I dt Z 0 0 1 2 004 10.0 points = (at − bt + c)dt The drift velocity of free electrons in a cop- 0 per wire is 6 mm/s. 1 = a t − bt + ct Find the electric ﬁeld in the conductor. 3 1 2 1 Copper’s resistivity is 1.61 × 108 Ω · m and 0 martinez (m6688) – Electric Current – satyal – (14442016S) 2 the free electron density is 8.44 × 1/m . What is the resistance of the wire? −19 The fundamental charge is 1.602 × 10 C. Correct answer: 2.11718 Ω. Correct answer: 1.30612 N/C. Explanation: Explanation: −8 Let : ρ = 3.92807 × 10 Ω · m, Let : vd= 6 mm/s = 0.006 m/s, ℓ = 5.8241 m, and n = 8.44 × 1028/m , −8 r = 0.18546 mm = 0.00018546 m. ρ = 1.61 × 10 Ω · m, and q = 1.602 × 10−19 C. From Ohm’s law, R = ρ ℓ = ρℓ E A π r2 J = σ E = = nq V d −8 ρ = 3.92807 × 10 Ω · m (5.8241 m) E = nq v d π (0.00018546 m)2 28 3 −19 = (8.44 × 10 /m )(1.602 × 10 C) = 2.11718 Ω . × (0.006 m/s)(1.61 × 10−8 Ω · m) = 1.30612 N/C . 007 10.0 points A wire with a circular cross section and a re- sistance R is lengthened to 3.47 times its orig- 005 10.0 points What is drift velocity? inal length by pulling it through a small hole. The total volume of the wire is unchanged. Find the resistance of the wire after it is 1. The speed of an electric ﬁeld stretched. 2. The highest speed of an electron in a Correct answer: 12.0409R. metal Explanation: 3. The average speed of electrons in a con- ductor in an electric ﬁeld correct ′ Let : ℓ = 3.47 ℓ. 4. The lowest speed of an electron in a metal A = V ∝ 1 when V = V , so ℓ ℓ 5. The average speed of atoms in a liquid A ℓ′ N ℓ ′ = = = N . Explanation: A ℓ ℓ When an electric ﬁeld is applied to a con- ductor, the electrons continue their random ρℓ ℓ ′ R = A ∝ A when ρ = ρ , so motions while simultaneously being nudged by this ﬁeld. The collisions interrupt the mo- ′ ′ tion of the electrons along the ﬁeld lines. The R = ℓ A = N ℓ N = N = (3.47) 2 average speed at which they migrate along a R ℓ A ′ ℓ ′ wire is known as drift velocity. R = 12.0409R . 006 10.0 points A wire is made of a material with a resis- 008 (part 1 of 2) 10.0 points tivity of 3.92807 × 10 Ω · m. It has length The potential diﬀerence in a simple circuit is 5.8241 m and diameter 0.37092 mm. 4 V and the resistance is 33 Ω. martinez (m6688) – Electric Current – satyal – (14442016S) 3 What current ﬂows in the circuit? From Ohm’s Law, ∆V = I R. The resis- tances are the same, so Correct answer: 0.121212 A. ∆V 1 ∆V 2 Explanation: = I1 I2 ∆V 2 1 (82 V)(0.4 A) Let : V = 4 V and 2 = = ∆V 1 431 V R = 33 Ω. = 0.0761021 A . The current is I = V = 4 V = 0.121212 A . 011 (part 2 of 2) 10.0 points R 33 Ω What is the current in the resistor when the operating potential diﬀerence is 472 V? 009 (part 2 of 2) 10.0 points How many electrons pass a given point in the Correct answer: 0.438051 A. circuit in 6 min? The fundamental charge is 1.602 × 10−19C. Explanation: 20 Let : ∆V = 472 V. Correct answer: 2.72387 × 10 . 3 ∆V 1 431 V Explanation: R = = = 1077.5 Ω, I1 0.4 A so the current is Let : I = 0.121212 A, t = 6 min = 360 s, and ∆V 3 472 V e = 1.602 × 10−19C. I3= = = 0.438051 A . R 1077.5 Ω I = q and the total charge for n electrons t is 012 10.0 points A silver wire has a resistance of 6 Ω at 20 C. q = ne ◦ What resistance does it have at 41 C? n = q = I t= (0.121212 A)(360 s) Use a temperature coeﬃcient of resistivity of e e 1.602 × 10−19 C 0.0038( C)−1 . Neglect any change in length = 2.72387 × 1020 . or cross-sectional area resulting from the tem- perature change. 010 (part 1 of 2) 10.0 points Correct answer: 6.4788 Ω. The current in a certain resistor is 0.40 A when it is connected to a potential diﬀerence Explanation: of 431 V. Let : R0= 6 Ω, What is the current in this same resistor if ◦ the operating potential diﬀerence is 82.0 V? T0= 20 C, T1= 41 C, and Correct answer: 0.0761021 A. α = 0.0038( C)−1 . Explanation: The resistance at 41 C is R = R 01 + α∆T) Let : 1 = 0.40 A, = 6 Ω 1 + 0.0038( C) −1 (21 C) ∆V 1 431 V, and ∆V = 82.0 V. = 6.4788 Ω . 2 martinez (m6688) – Electric Current – satyal – (14442016S) 4 A 12 V automobile battery is connected to an 013 10.0 points electric starter motor. The current through An electron moving in a wire collides again the motor is 208 A. and again with atoms and travels an average How much power does the motor dissipate? distance between collisions that is called the mean free path. Correct answer: 2496 W. For a given conductor, what can you do to Explanation: lengthen the mean free path? Power is 1. Cool the conductor correct P = I V = (208 A)(12 V) = 2496 W . 2. Apply a large voltage across the conduc- tor 016 (part 2 of 2) 10.0 points 3. Heat the conductor How much energy does the battery deliver to the motor during a 25 s period? 4. Place the conductor in a vacuum Explanation: Correct answer: 62400 J. For all materials, the application of heat Explanation: imposes more molecular chaos and shortens the path even more, increasing resistance in most materials; to lengthen the path, simply Let : V = 12 V, cool the material. I = 208 A, and t = 25 s. 014 10.0 points A particular type of automobile storage bat- Energy is tery is characterized as “318 - Ampere - hour, E = P · T = (I V )t 8.6 V.” What total energy can the battery deliver? = (2496 W)(25 s) 6 = 62400 J . Correct answer: 9.84528 × 10 J. Explanation: 017 (part 1 of 4) 10.0 points An object with a resistance of 43 Ω has 130 V Let : I = 318 A · hr and applied to it. T How much electric current is going through V = 8.6 V. this object? E Correct answer: 3.02326 amps. P = t Explanation: E = P t = I T Let : R = 43 Ω and = (318 A · hr)(8.6 V) 3600 s V = 130 V. 1 hr 6 Applying Ohm’s law, = 9.84528 × 10 J . V = I R I = V = 130 V = 3.02326 A . 015 (part 1 of 2) 10.0 points R 43 Ω martinez (m6688) – Electric Current – satyal – (14442016S) 5 If electrical energy costs 10 cents per 018 (part 2 of 4) 10.0 points kilowatt-hour, how much does it cost to oper- How much power is being produced by the ater the coﬀeepot for 2 hours? object? 1. 8.0 cents Correct answer: 393.023 W. Explanation: 2. 9.6 cents correct 3. 2.4 cents P = I V = (3.02326A)(130V) = 393.023 W . 4. 4.8 cents 5. 16 cents 019 (part 3 of 4) 10.0 points How much electric energy will be needed to Explanation: operate this object for 22 h? The power that the coﬀeepot consumes is Correct answer: 8.64651 kWh. P = IU = (4.0A)(120V) = 480W, Explanation: so the cost to operate the coﬀeepot for 2 hours is Let : t = 22 h. 10cents cost = (480W)(2hrs) kW · h E 10cents P = = (960W · h) t kW · h E = P t 10cents 1 kW = (0.96kW · h) = (393.023 W)(22 h) 1000 W kW · h = 8.64651 kWh . = 9.6cents . 020 (part 4 of 4) 10.0 points What is the cost of operating this object for 22 h at a rate of 6.5 cents/kWh? Correct answer: $0.562023. Explanation: Let : R = 6.5 cents/kWh = $0.065 /kWh. C = ( $0.065/kWh)(8.64651kWh) = $0.562023 . 021 10.0 points A certain coﬀeepot draws 4.0 A of current when it is operated on 120 V household lines.

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I made $350 in just two days after posting my first study guide."

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.