Dr. West, Montclair State University
Physics for Scientists and Engineers by Serway and Jewett, 8th Edition, 2010
Please answer the problems using WebAssign (webassign.net, smithp3, montclair, smithp3)
using YOUR MSU email username instead of smithp3.
- For Monday, September 12 (4 points)
Chapter 1:
Problems 1, 9, 19
- For Monday, September 19 (13 points)
Chapter 2: Problems 1, 9, 17, 37
- For Monday, September 26 (9 points)
Chapter 3: Problems 1, 7, 21, 29
1. For r = 5.50 m, and Theta = 240 degrees, x = -2.75 meters, and y = -4.76 meters.
7. For d = 100 meters and theta = 35.0 degrees, width = 70.0 meters.
21. For 75 m North, 250 m East, 125 m at an angle of 30 degrees north of East, and 150 m South,
the resultant displacement is 358 m at 2.0 degrees south of East.
29. For Force (1) = 120 N at a direction of 60.0 degrees, and Force (2) = 80 N at a direction of 75 degrees,
the single equivalent force is -39.3 N i and -181 N j.
- For Thursday, September 29 (10 points)
Vector Practice Sheet
- For Monday, October 3 (14 points)
Chapter 4. Problems 1, 5, 9, 19
1. For driving speed 20 m/s south for 3.00 minutes, then west at 25.0 m/s for 2.0 minutes, then northwest at 30.0 m/s for 1 minute.
a = 4.87 km at 209 degrees from East.
b = 23.3 m/s
c = 13.5 m/s at 209 degrees
2. If r = 3.00 i - 6.00 t^2 j, then v = -12 t j, and a = -12 j, and at 1.00 sec r = 3 i - 6 j, and v = -12 j
3. The local bar has a counter 1.22 m high, and the beer mug hits the floor 1.40 m from the base. The v = 2.81 m/s and the direction is 60.2 degrees.
4. The playground is 7.0 m above the street, wall height = 1.0 m, kicked at theta = 53 degrees, distance = 24 m, t = 2.2 seconds to be above the wall.
speed = 18.1 m/s, height clearing wall = 1.13 m, distance = 2.79 m
This page is http://www.csam.montclair.edu/~west/phys191/phys191hw.html