Ang iven m, = 93 Jm そつさてい→k22-C Tabing moment mig L45.scm at poiut O E50cm -2 2 migx 225x10 = mgxo.5X10 -3 23メ1b メ22.5 = mx0.5 -3 m= 1035XI0 m= 1.025kg mass of mele+ tick is 1.03S kg

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter10: Rotational Motion
Section: Chapter Questions
Problem 29P
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this is solution true or false

Ans given
m, = 23
に23gom
ぐっさてい→k22:<5
Taking
mig
-45.scn
moment
at pošut o
-2
mgxo.sxio
(ID
メ
ー3
23 X10 x
X 22.5 =
mx 0.5
-3
1035X10
m= 1:025 kg
mass of meler Stick i8 1.035
meter Stick i81.035 kg
Transcribed Image Text:Ans given m, = 23 に23gom ぐっさてい→k22:<5 Taking mig -45.scn moment at pošut o -2 mgxo.sxio (ID メ ー3 23 X10 x X 22.5 = mx 0.5 -3 1035X10 m= 1:025 kg mass of meler Stick i8 1.035 meter Stick i81.035 kg
42/306
In a classroom demonstration, a meter stick is
balanced at its midpoint on a narrow support.
When a 23.0 g candy piece is
placed at the 25.0 cm mark, the support must
be moved to the 45.5 cm mark to rebalance
the stick.What is the stick's mass?
Transcribed Image Text:42/306 In a classroom demonstration, a meter stick is balanced at its midpoint on a narrow support. When a 23.0 g candy piece is placed at the 25.0 cm mark, the support must be moved to the 45.5 cm mark to rebalance the stick.What is the stick's mass?
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