Asked by Joy
d1 = 2.53 cm +/- .05 cm
d2 = 1.753 m +/- .001 m
0 = 23.5 degrees +/- .5 degrees
v1 = 1.55 m/s +/- .15 m/s
Using the measured quantities above, calculate the following. Express the uncertainty calculated value.
1. d3 = 4 (d1 + d2)
d2 = 1.753 m +/- .001 m
0 = 23.5 degrees +/- .5 degrees
v1 = 1.55 m/s +/- .15 m/s
Using the measured quantities above, calculate the following. Express the uncertainty calculated value.
1. d3 = 4 (d1 + d2)
Answers
Answered by
Count Iblis
If f(x1,x2) is a function of x1 and x2 which have independent uncertainties sigma1 and sigma2, then the uncertainty in f(x1,x2) is:
sigmaf = sqrt[sigmaf1^2 +sigmaf2^2]
where sigmaf1 and sigmaf2 are the contributions to the uncertainty in f coming from x1 and x2:
sigmaf1 = f(x1+sigma1,x2) - f(x1,x2)
sigmaf2 = f(x1,x2+sigma2) - f(x1,x2)
Clearly, the uncertainty in d1 makes a contribution to the uncertainty in d3 of:
4*0.05 cm = 0.2 cm
while the uncertainty in d2 makes a contribution to the uncertainty in d3 of:
4*0.1 cm = 0.4 cm
The uncertainty in d3 is then:
sqrt[(0.2 cm)^2 + (0.4 cm)^2] =
0.45 cm
sigmaf = sqrt[sigmaf1^2 +sigmaf2^2]
where sigmaf1 and sigmaf2 are the contributions to the uncertainty in f coming from x1 and x2:
sigmaf1 = f(x1+sigma1,x2) - f(x1,x2)
sigmaf2 = f(x1,x2+sigma2) - f(x1,x2)
Clearly, the uncertainty in d1 makes a contribution to the uncertainty in d3 of:
4*0.05 cm = 0.2 cm
while the uncertainty in d2 makes a contribution to the uncertainty in d3 of:
4*0.1 cm = 0.4 cm
The uncertainty in d3 is then:
sqrt[(0.2 cm)^2 + (0.4 cm)^2] =
0.45 cm
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