physics homework 2

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• Dr. Michelle Tomasik
• Prof. Krishna Rajagopal
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Physics ii: electricity and magnetism, course description.

Electricity and magnetism dominate much of the world around us – from the most fundamental processes in nature to cutting-edge electronic devices. Electric and magnetic fields arise from charged particles. Charged particles also feel forces in electric and magnetic fields. Maxwell’s equations, in addition to describing …

Electricity and magnetism dominate much of the world around us – from the most fundamental processes in nature to cutting-edge electronic devices. Electric and magnetic fields arise from charged particles. Charged particles also feel forces in electric and magnetic fields. Maxwell’s equations, in addition to describing this behavior, also describe electromagnetic radiation. 

The three-course series comprises:

8.02.1x : Electrostatics 8.02.2x : Magnetic Fields and Forces 8.02.3x : Maxwell’s Equations

This course was organized as a three-part series on MITx by MIT’s Department of Physics and is now archived on the Open Learning Library , which is free to use. You have the option to sign up and enroll in each module if you want to track your progress, or you can view and use all the materials without enrolling.

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Problems & Exercises

(c) + 7 m + 7 m

a. 8  m + 2  m + 3  m = 13  m ¯ 8  m + 2  m + 3  m = 13  m ¯

b. 9  m ¯ 9  m ¯

c. Δ x = 11  m − 2  m = 9  m ¯ Δ x = 11  m − 2  m = 9  m ¯

(a) 3 . 0 × 10 4 m/s 3 . 0 × 10 4 m/s

2 × 10 7 years 2 × 10 7 years

34 . 689 m/s = 124 . 88 km/h 34 . 689 m/s = 124 . 88 km/h

(a) 40 . 0 km/h 40 . 0 km/h

(b) 34.3 km/h, 25º S of E . 25º S of E .

(c) average speed = 3.20 km/h, v - = 0. average speed = 3.20 km/h, v - = 0.

(a) 6 . 61 × 10 15 rev/s 6 . 61 × 10 15 rev/s

4 . 29 m/s 2 4 . 29 m/s 2

(a) 1 . 43 s 1 . 43 s

(b) − 2 . 50 m/s 2 − 2 . 50 m/s 2

(a) 10 . 8 m/s 10 . 8 m/s

38.9 m/s (about 87 miles per hour)

(a) 16 . 5 s 16 . 5 s

(b) 13 . 5 s 13 . 5 s

(c) − 2 . 68 m/s 2 − 2 . 68 m/s 2

(a) 20 . 0 m 20 . 0 m

(b) − 1 . 00 m/s − 1 . 00 m/s

(c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at 2 . 00 m/s 2 2 . 00 m/s 2 , then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.

0 . 799 m 0 . 799 m

(a) 28 . 0 m/s 28 . 0 m/s

(b) 50 . 9 s 50 . 9 s

(c) 7.68 km to accelerate and 713 m to decelerate

(a) 51 . 4 m 51 . 4 m

(b) 17 . 1 s 17 . 1 s

(a) − 80 . 4 m/s 2 − 80 . 4 m/s 2

(b) 9 . 33 × 10 − 2 s 9 . 33 × 10 − 2 s

(a) 7 . 7 m/s 7 . 7 m/s

(b) − 15 × 10 2 m/s 2 − 15 × 10 2 m/s 2 . This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high!

(a) 32 . 6 m/s 2 32 . 6 m/s 2

(b) 162 m/s 162 m/s

(c) v > v max v > v max , because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32 . 6 m/s 2 32 . 6 m/s 2 during the last few meters, but substantially less, and the final velocity would be less than 162 m/s.

(a) v = 12 . 2 m/s v = 12 . 2 m/s ; a = 4 . 07 m/s 2 a = 4 . 07 m/s 2

(b) v = 11 . 2 m/s v = 11 . 2 m/s

(a) y 1 = 6 . 28 m y 1 = 6 . 28 m ; v 1 = 10 . 1 m/s v 1 = 10 . 1 m/s

(b) y 2 = 10 . 1 m y 2 = 10 . 1 m ; v 2 = 5 . 20 m/s v 2 = 5 . 20 m/s

(c) y 3 = 11 . 5 m y 3 = 11 . 5 m ; v 3 = 0 .300 m/s v 3 = 0 .300 m/s

(d) y 4 = 10 .4 m y 4 = 10 .4 m ; v 4 = − 4 .60 m/s v 4 = − 4 .60 m/s

v 0 = 4 . 95 m/s v 0 = 4 . 95 m/s

(a) a = − 9 . 80 m/s 2 a = − 9 . 80 m/s 2 ; v 0 = 13 . 0 m/s v 0 = 13 . 0 m/s ; y 0 = 0 m y 0 = 0 m

(b) v = 0 m/s v = 0 m/s . Unknown is distance y y to top of trajectory, where velocity is zero. Use equation v 2 = v 0 2 + 2 a y − y 0 v 2 = v 0 2 + 2 a y − y 0 because it contains all known values except for y y , so we can solve for y y . Solving for y y gives

Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.

(c) 2 . 65 s 2 . 65 s

(b) 0.717 s

(a) -70.0 m/s (downward)

(a) 19 . 6 m 19 . 6 m

(b) 18 . 5 m 18 . 5 m

(b) 262 m, -29.2 m/s

(a) 115 m/s 115 m/s

(b) 5 . 0 m/s 2 5 . 0 m/s 2

(c) 3 m/s 2 3 m/s 2

(a) Car A is traveling faster at the checkpoint because it must go past the speed of car B to reach the same distance.

(b) i. Yes, the equation is consistent with the answer because the speed of car A is only a constant away from the correct answer. ii. Yes, the equation makes sense because V = 2 V 0 V = 2 V 0 .

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• Authors: Paul Peter Urone, Roger Hinrichs
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• Book title: College Physics 2e
• Publication date: Jul 13, 2022
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
• Section URL: https://openstax.org/books/college-physics-2e/pages/chapter-2

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