From the slope of the graph, find the mass of the system and compare it to the expected value of 4kg using % Difference. This involves a derivation – show how you get the mass from the slope. Experimental Acceleration vs Accelerating Force8y = 1.0107x.esi711234567Accelerating Forceco6.st2.Experimental Acceleration Eq 1.Eq 2.EFEq 3.ΣεF = Whanging= m19m19at?1аехресtedAy =MM(1kg)(9.8)/1kg+3kg =2.45 m/s^2(2kg)(9.8)/2kg+2kg =4.90 m/s^21kg(9.8) = 9.8 N2(2.02m)/(1.28s)^2= 2.47 m/s^22(2.02m)/(0.91s)^2= 4.89 m/s^2Run 1Run 22kg(9.8) = 19.6 N%3!Run 33kg(9.8) = 29.4 N(3kg)(9.8)/3kg+1kg =7.35 m/s^22(2.02m)/(0.74s)^2= 7.38 m/s^2%3D%3!

Answered: Experimental Acceleration vs…

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter34: Frontiers Of Physics

Section: Chapter Questions

Problem 22CQ: Must a complex system be adaptive to be of interest in the field of complexity? Give an example to...

See similar textbooks

Related questions

Question

From the slope of the graph, find the mass of the system and compare it to the expected value of 4kg using % Difference. This involves a derivation – show how you get the mass from the slope.

Experimental Acceleration vs Accelerating Force
8
y = 1.0107x.esi
7
1
1
2
3
4
5
6
7
Accelerating Force
co
6.
st
2.
Experimental Acceleration

Eq 1.
Eq 2.
EF
Eq 3.
Σε
F = Whanging
= m19
m19
at?
1
аехресted
Ay =
M
M
(1kg)(9.8)/1kg+3kg =
2.45 m/s^2
(2kg)(9.8)/2kg+2kg =
4.90 m/s^2
1kg(9.8) = 9.8 N
2(2.02m)/(1.28s)^2
= 2.47 m/s^2
2(2.02m)/(0.91s)^2
= 4.89 m/s^2
Run 1
Run 2
2kg(9.8) = 19.6 N
%3!
Run 3
3kg(9.8) = 29.4 N
(3kg)(9.8)/3kg+1kg =
7.35 m/s^2
2(2.02m)/(0.74s)^2
= 7.38 m/s^2
%3D
%3!

Expert Solution

Step 1

We are given the graph of acceleration vs accelerating force and the graph is a straight line as follows :

$y = 1.0107 x^{0.9951}$

Determine:

the mass of the system by using slope of the graph

Trending now

This is a popular solution!

Step by step

Solved in 5 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions