For item 25:TABLE 1. Determining the Weight of Pans12Actual value of pan 1, P₁ = 25 gramsActual value of pan 2, P₂ = 24.8 gramsL₁L₂For item 25:Ⓒ 21.2220.2223.22Ⓒ24.2228.22TrialW₁ = 10 gW₂ = 5 gW₁ =15 gW₂ =25 g14.9 cm12.6 cmchoices for number 2521 cm19.8 cmL316.2 cm23.7 cmL₁13.4 cm 25.)11.7 cmFor item 26:26.2228.3723.37P₁ (computed)C25.3724.22gP₂ (computed)24.28 g24.87 g 26.)choices for number 26g

From torque equilibrium , P1 =(P2+W2)L4L3 P2 =(P1+W1)L1L2

Actual value of pan 1, P₁ = 25 grams Actual value of pan 2, P₂ = 24.8 grams L₁ L₂ Trial W₁ =10 g W₂ = 5 g 15 g 14.9 cm 21 cm L3 16.2 cm LA P₁ (computed) 13.4 cm 25.) g P₂ (computed) 24.28 g

Actual value of pan 1, P₁ = 25 grams Actual value of pan 2, P₂ = 24.8 grams L₁ L₂ Trial W₁ =10 g W₂ = 5 g 15 g 14.9 cm 21 cm L3 16.2 cm LA P₁ (computed) 13.4 cm 25.) g P₂ (computed) 24.28 g

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter1: Introduction: The Nature Of Science And Physics

Section: Chapter Questions

Problem 33PE: Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice...

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For item 25:
TABLE 1. Determining the Weight of Pans
1
2
Actual value of pan 1, P₁ = 25 grams
Actual value of pan 2, P₂ = 24.8 grams
L₁
L₂
For item 25:
Ⓒ 21.22
20.22
23.22
Ⓒ24.22
28.22
Trial
W₁ = 10 g
W₂ = 5 g
W₁ =15 g
W₂ =25 g
14.9 cm
12.6 cm
choices for number 25
21 cm
19.8 cm
L3
16.2 cm
23.7 cm
L₁
13.4 cm 25.)
11.7 cm
For item 26:
26.22
28.37
23.37
P₁ (computed)
C25.37
24.22
g
P₂ (computed)
24.28 g
24.87 g 26.)
choices for number 26
g

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