Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 2, Problem 4P
An athlete leaves one end of a pool of length L at t = 0 and arrives at the other end at time t1. She swims back and arrives at the starting position at time t2. If she is swimming initially in the positive x direction, determine her average velocities symbolically in (a) the first half of the swim, (b) the second half of the swim, and (c) the round trip. (d) What is her average speed for the round trip?
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A block of mass m = 7.00 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Assume h₂ = 7.80 m.)
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(a) Determine the block's speed at points ® and
point B
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m/s
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(b) Determine the net work done by the gravitational force on the block as it moves from point
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A 1.10 x 10²-g particle is released from rest at point A on the inside of a smooth hemispherical bowl of radius R
R
B
2R/3
(a) Calculate its gravitational potential energy at A relative to B.
]
(b) Calculate its kinetic energy at B.
]
(c) Calculate its speed at B.
m/s
(d) Calculate its potential energy at C relative to B.
J
(e) Calculate its kinetic energy at C.
]
=
26.5 cm (figure below).
Report on the percentage errors (with uncertainty) between the value of 'k' from the F vs displacement plot and each of the values of 'k' from the period measurements. Please comment on the goodness of the results.
Value of k = Spring constant k = 50.00 N/m
Each of the values of k from period measurements:
Six Measurements of time for 5 osccilations: t1 = 7.76s, t2=8.00s, t3=7.40s, t4=7.00s, t5=6.90s, t6=7.10s
(t1-tavg)^2 = (7.76-7.36)^2 = 0.16%(t2-tavg)^2 =(8.00-7.36)^2 = 0.4096%(t3-tavg)^2 =(7.40-7.36)^2 = 0.0016%(t4-tavg)^2 =(7.00-7.36)^2 = 0.1296%(t5-tavg)^2 =(6.90-7.36)^2 = 0.2116%(t6-tavg)^2 =(7.10-7.36)^2 = 0.0676
Chapter 2 Solutions
Physics for Scientists and Engineers with Modern Physics
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