these are the questions thank you in advance Latest e e ioeF FORCE vs TIME (use
ID: 1770669 • Letter: T
Question
these are the questions
thank you in advance
Latest e e ioeF FORCE vs TIME (use Asin(Bt+C)+D for CurveFit) 1.99 003515 0.00 19790.01598 0.01 2 0.05 5.6 0.68 4.52 0.67 4.5 1.9280005673 0 4 015 6 5 0.20 6 0.66 4.58 6 0.25 6.81 0.65 4.61 810.35 4 101-045 1.876.000833 0.001 7 0.30 7.19 0.64 4651.8260007814 0.001 08239 0.001 1.750.001103 0.001 0.40 8 0.62 4.71 4.7 0 0.45 8.4 B: 4789 0001782 C: 4.752 +-0.005341 D: 349 +/-0 003300 1.721.001762 0.001 09997 RMSE: 0 03377 N Time (s) record B and enter the value in the column 4- co(B) record the ermor on B and enter it in the coulmn 6 - r(w) arter t in the Linear Fit for. Latest | TSquared T2 = mx+b 19b (Y-Intercept): 0 07457 +0.04472 s 09972 Force 7.04 N 2 1.8 0.65 Mass (kg) 0.7Explanation / Answer
from the given graphs
slope = 2.739 +- 0.04472 s^2/kg of the graph T^2 vs m
a. hence slope = 2.739 s^2/kg
error = 0.04472 s^2/kg
b. now, spring constant = k
we know that for a spring mass system
T = 2*pi*sqroot(m/k)
hence
T^2 = 4*pi^2*m/k
T^2/m = 4*pi^2/k
hence
2.739 = 4*pi*pi/k
k = 14.41344 N/m
c. error in k = d(k)
from the given formula
d(k) = 4*pi*pi/slope^2 * d(slope)
or
d(k)/k = d(slope)/slope
d(k) = 14.41344*0.04472/2.793
d(k) = 0.246261 N/m
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