7) A bob attached to a cord is moved to the right where its vertical position is 1.05 cm abo equilibrium position and is then given an initial speed of 0.6 m/s. What are the values of the max speed and maximum height reached by the bob? (Take g = 9.8 m/s²) (a) hmax (b) hmax = 2.89 cm; vmax = 0.75 m/s (c) hmax %3D 1.87 cm; vmax = 3.44 m/s %3D = 1.87 cm; Vmax = 0.75 m/s (d)hmax = 2.89 cm; vmax = 3.44 m/s

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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(c) +r/2
(d)n
7) A bob attached to a cord is moved to the right where its vertical position is 1.05 cm abov
equilibrium position and is then given an initial speed of 0.6 m/s. What are the values of the maxin
speed and maximum height reached by the beb? (Take g = 9.8 m/s?)
%3D
(a) hmax = 1.87 cm; vmax
(b)hmax
(c) hmax
(d)hmax =
3.44 m/s
0.75 m/s
= 1.87 cm; vmāx = 0.75 m/s
2.89 cm; vmax = 3.44 m/s
2.89
qm; Vmax =
%3D
8) A simple pendulum of length L and mass M has frequency f. To increase its frequency te 2f:
(a) decrease its length to L/4
(b) decrease its length to L/2
(c) increase its length to 2L
(d)increase its length to 4L
9) A simple pendulum of length L is set to oscillate in simple harmonic motion. When its potential energy
is one-half its total mechanical energy, U = E/2, then which of the following is true about its velocity:
(a) v = tvmax/2
(b)v = tax/4
(c) v = +vmax V2/2
(d)v = tvnax V3/2
at to ose
in simpl
otion The
101
A simple nandulum.of lenoth /
Transcribed Image Text:(c) +r/2 (d)n 7) A bob attached to a cord is moved to the right where its vertical position is 1.05 cm abov equilibrium position and is then given an initial speed of 0.6 m/s. What are the values of the maxin speed and maximum height reached by the beb? (Take g = 9.8 m/s?) %3D (a) hmax = 1.87 cm; vmax (b)hmax (c) hmax (d)hmax = 3.44 m/s 0.75 m/s = 1.87 cm; vmāx = 0.75 m/s 2.89 cm; vmax = 3.44 m/s 2.89 qm; Vmax = %3D 8) A simple pendulum of length L and mass M has frequency f. To increase its frequency te 2f: (a) decrease its length to L/4 (b) decrease its length to L/2 (c) increase its length to 2L (d)increase its length to 4L 9) A simple pendulum of length L is set to oscillate in simple harmonic motion. When its potential energy is one-half its total mechanical energy, U = E/2, then which of the following is true about its velocity: (a) v = tvmax/2 (b)v = tax/4 (c) v = +vmax V2/2 (d)v = tvnax V3/2 at to ose in simpl otion The 101 A simple nandulum.of lenoth /
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