200 216.2 0.2165 2.1271 250 8 266.55 02. 266.5 2.6/17 Average force constant Kaug Force constant Ktren from the slope of the trendline in the graph % relative difference between Kavg and Ktren Mhi Mass of the weight hanger 1. Xo: Position

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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the
2
5
13.0
2.5
2.6
1.5-
1.0
60.5
yo
‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒
Please show here the calculation of the slope.
Slope =
0.00 Op
Figure.
32.5n/m
OUL
0.03
0.04 0.05 006 007 0.08
Transcribed Image Text:the 2 5 13.0 2.5 2.6 1.5- 1.0 60.5 yo ‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒ Please show here the calculation of the slope. Slope = 0.00 Op Figure. 32.5n/m OUL 0.03 0.04 0.05 006 007 0.08
Figure 4. How to correctly measure the position of the bottom of the spring when the
weight hanger carries the required added masses.
M₁ (8) = 6₁54
Msw
(g)
Table # 1: Determination of the spring constant K
g= 9.80 m/s²
xo (cm) = 2
MT= Mh+ Msw
MT
MT
(kg)
6.6.5, 0.066.5
116:52 0.11 6.5
F =W=Mrg
F
(N)
0
0.6517
1.1417
xf
(cm)
11.7
3.3
4.9
X = Xf- Xo
X
(cm)
X
(m)
0
50
100
150
200
166.5 0.766.5 1.63/7
216.30.2165 2.1217
266.55 (2.2665 2.6/17
Average force constant Kavg
250
8
Force constant Ktren from the slope of the trendline in the graph
% relative difference between Kavg and Ktren
Mh: Mass of the weight hanger
Msw: Mass of added slotted weights
Mr: Total Mass
W: Weight of hanger plus slotted weights
K =
F(N)
x(m)
K
(N/m)
0.2
0.1
65√7
2.5ch 0.025 45:068
You 0.09 40,7925
0:055 38 57636364
7.2 0.072 36.973 6 UI!
55
35.7829 n/m
Xo: Position of the spring's bottom (unstretched)
xf: Position of the spring's bottom (stretched)
x: Elongation of the spring
K: Force constant of the spring
25
2.
0.0.
C
Transcribed Image Text:Figure 4. How to correctly measure the position of the bottom of the spring when the weight hanger carries the required added masses. M₁ (8) = 6₁54 Msw (g) Table # 1: Determination of the spring constant K g= 9.80 m/s² xo (cm) = 2 MT= Mh+ Msw MT MT (kg) 6.6.5, 0.066.5 116:52 0.11 6.5 F =W=Mrg F (N) 0 0.6517 1.1417 xf (cm) 11.7 3.3 4.9 X = Xf- Xo X (cm) X (m) 0 50 100 150 200 166.5 0.766.5 1.63/7 216.30.2165 2.1217 266.55 (2.2665 2.6/17 Average force constant Kavg 250 8 Force constant Ktren from the slope of the trendline in the graph % relative difference between Kavg and Ktren Mh: Mass of the weight hanger Msw: Mass of added slotted weights Mr: Total Mass W: Weight of hanger plus slotted weights K = F(N) x(m) K (N/m) 0.2 0.1 65√7 2.5ch 0.025 45:068 You 0.09 40,7925 0:055 38 57636364 7.2 0.072 36.973 6 UI! 55 35.7829 n/m Xo: Position of the spring's bottom (unstretched) xf: Position of the spring's bottom (stretched) x: Elongation of the spring K: Force constant of the spring 25 2. 0.0. C
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