A uniform 1.5-kg stick has acenter of mass located at 30cm from one end. If it ispivoted about that end, itocillates with a period of 2 s.Calculate its moment of inertiaabout this end?O 0.45 kg.m2O 44.73 kg.m2O 0.03 kg.m2O 2.80 kg.m2 A geophysicist uses a smallpendulum of length 0.171 m tocalculate the acceleration due3.iftto gravity g. if the pendulumhas 70 complete swings in atime of 50.0 s. What is thevalue of g in the geophysicist'slocation?O 3.44 m/s22.26 m/s2O 13.22 m/s2O 9.8 m/s2

Given that,m=1.5 kgd=0.30 mt=2 s

A uniform 1.5-kg stick has a center of mass located at 30 cm from one end. If it is pivoted about that end, it ocillates with a period of 2 s. Calculate its moment of inertia about this end? O 0.45 kg.m2 O 4.73 kg.m2 O 0.03 kg.m2 O 2.80 kg.m2

A uniform 1.5-kg stick has a center of mass located at 30 cm from one end. If it is pivoted about that end, it ocillates with a period of 2 s. Calculate its moment of inertia about this end? O 0.45 kg.m2 O 4.73 kg.m2 O 0.03 kg.m2 O 2.80 kg.m2

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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Question

A uniform 1.5-kg stick has a
center of mass located at 30
cm from one end. If it is
pivoted about that end, it
ocillates with a period of 2 s.
Calculate its moment of inertia
about this end?
O 0.45 kg.m2
O 44.73 kg.m2
O 0.03 kg.m2
O 2.80 kg.m2

A geophysicist uses a small
pendulum of length 0.171 m to
calculate the acceleration due
3.ift
to gravity g. if the pendulum
has 70 complete swings in a
time of 50.0 s. What is the
value of g in the geophysicist's
location?
O 3.44 m/s2
2.26 m/s2
O 13.22 m/s2
O 9.8 m/s2

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The equations used to describe the physical pendulum will lead us directly to an answer. Part 3 of 3 - Analyze We are givenf = 0.685 Hz, d 0.510 m, and m = 2.10 kg. We have the following equation for the period. T = 2n V mgd This gives 4721 T2 = mgd and, solving for the moment of inertia, we have the following. T?mgd mgd %3D kg )(9.80 m/s?) m Hz kg - m2 Submit Skip (you cannot come back) Need Help? Read it O Show My Work (Optional) ?