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Physics (5th Edition)
5th Edition
ISBN: 9780134051802
Author: Walker
Publisher: PEARSON
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Textbook Question
Chapter 1, Problem 55GP
Acceleration is related to velocity and time by the following expression a = v2t2. Find the powers p and q that make this equation dimensonaiy consistent.
Expert Solution & Answer
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Students have asked these similar questions
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.)
a
m
ha
3.20 m
2.00 m
i
(a) Determine the block's speed at points ® and
point B
©.
m/s
m/s
point
(b) Determine the net work done by the gravitational force on the block as it moves from point
J
A
to point
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 1 Solutions
Physics (5th Edition)
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