is my answer and process correct in the pic? what is the answer to this?Consider the percentage difference between the and for the first two parts of the laboratory when known forces are balanced. A difference of 0.5% or less is excellent, a difference of 1.0% or less is good, and a difference of 2% or less is acceptable. Based on these criteria, describe your results for the first two parts of the laboratory and defend your statement. Data and Calculations Table 3Actual mass of meter stick, mo = 0.0722 kg!!Mass (kg)Position (m)Lever arm (m) (m.)exp% Errorm1 = 0.200X1 = 0.100di = 0.40m2 = 0.050X2 = 0.80d2 = 0.300.475757.895%Xo = 0.3do= 0.20Lever armD1= 0.5 - 0.1 = 0.40mD2= 0.5 – 0.80 = 0.30mD3= 0.5 - 0.3 = 0.20m(mo)expm1(xp-x1)+m2(x2-xp) 0.200(0.50-0.100)+0.050(0.80-0.50)-0.475mo=-(х0—хр)(0.3-0.50)% Error/(-0.475)-0.0722/(100)= 757.895%0.0722

As per the guidelines, the percentage difference between experimental answer and theoretical answer…

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College Physics

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ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter8: Linear Momentum And Collisions

Section: Chapter Questions

Problem 6PE: The mass of Earth is 5.9721024 kg and its orbital radius is an average of 1.4961011 m. Calculate its...

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is my answer and process correct in the pic?

what is the answer to this?

Consider the percentage difference between the and for the first two parts of the laboratory when known forces are balanced. A difference of 0.5% or less is excellent, a difference of 1.0% or less is good, and a difference of 2% or less is acceptable. Based on these criteria, describe your results for the first two parts of the laboratory and defend your statement.

Data and Calculations Table 3
Actual mass of meter stick, mo = 0.0722 kg
!!
Mass (kg)
Position (m)
Lever arm (m) (m.)exp
% Error
m1 = 0.200
X1 = 0.100
di = 0.40
m2 = 0.050
X2 = 0.80
d2 = 0.30
0.475
757.895%
Xo = 0.3
do= 0.20
Lever arm
D1= 0.5 - 0.1 = 0.40m
D2= 0.5 – 0.80 = 0.30m
D3= 0.5 - 0.3 = 0.20m
(mo)exp
m1(xp-x1)+m2(x2-xp) 0.200(0.50-0.100)+0.050(0.80-0.50)
-0.475
mo=-
(х0—хр)
(0.3-0.50)
% Error
/(-0.475)-0.0722/
(100)= 757.895%
0.0722

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