error analysis for ipho contestants
TRANSCRIPT
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7/30/2019 Error Analysis for IPhO Contestants
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= (7.297 352 537 0.000000005) 103
1 = 1.1 0.5 kg/m3 2 = 1.5 0.4 kg/m3
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7/30/2019 Error Analysis for IPhO Contestants
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x = 02
a b b b/|a|
0
2 ek(xx0)2 k x0 = 2
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7/30/2019 Error Analysis for IPhO Contestants
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fa b
fa = s b = t
f = f(a, b) = a/ba a b b
ff(a a,b b)
f ff
f = max {f f, f f } .f
s = 10 1 cm t =5 0.2 s
q =ac
bsin(a) ln d,
a = 0.8 0.1 b = 2 0.3 c = 0.4 0.01d = 10 2
q = a
b
q
|q| =a
|a| +b
|b| .
a/|a| 1q = a/b
1
1 x 1 x
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7/30/2019 Error Analysis for IPhO Contestants
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x
1
q =a1 a2 aNb1 b2 bM
q = x + y q = x + yx y
q =
(x)2 + (y)2.
q
q=
a1a1
2+ +
aNaN
2
q = a1 ak/(ak+1 aN)
q q qq =
(q )2 + (q )2
L = 7.50 cm0.10 cm
0.05 cmL =
(0.10 cm)2 + (0.05 cm)2 = 0.12 cm
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7/30/2019 Error Analysis for IPhO Contestants
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m
h t
e =mgh
V It,
V Im h V I
t
U = RI
t1/2t
N(t) = N0 exp(t) = N0 exp ln 2
t1/2t
,
N0 t = 0 = ln 2/t1/2
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7/30/2019 Error Analysis for IPhO Contestants
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tA(t) 103
A(t) = |N(t)| = N0 exp(t).
f(x) = A
exp(
Bt)
ln A(t) = ln
N0et
= ln(N0) te
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7/30/2019 Error Analysis for IPhO Contestants
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t = 0
x x1, x2, . . . , xN
=
Ni=1(x xi)2
N.
N 1 NN 1 N
N N 1 N
x =N
.
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7/30/2019 Error Analysis for IPhO Contestants
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x Nxi xi1/x2i
x =
Ni=1 xi/x
2iN
i=1 1/x2i
x =1N
i=1 1/x2i
.
x x1 = 5 x2 = 8x3 = 10 x4 = 7 x5 = 6 x7 = 6 x8 = 22 x = 9 x10 = 7 x11 = 4x12 = 5
x q = q(x , y , z , . . . )
xq
y zq(x) q(x)
q(x + x) = q(x) + q(x)x.
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7/30/2019 Error Analysis for IPhO Contestants
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q(x) x x q
q = q(x + x) q(x) = q(x)x.
xx x
x2
x
y z . . .
q =
q
xx
2+
q
yy
2+
q
zz
2
q = q(x,y,z)
= 30 2 = 450 30 nm
d sin = k d k
q = x1 xN
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7/30/2019 Error Analysis for IPhO Contestants
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q = x2
q
q= 2
x
|x| .
q = x x
q
q=
x
x
2+
x
x
2=
2x
|x| .
q = x(ln x 1)x x = 1.00 0.07 q
1%a 0.01 a
0.9 % + 1
10.00 V 0.009 10.00 V+0.01 V
(0.09 + 0.01) V = 0.1 V
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7/30/2019 Error Analysis for IPhO Contestants
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U = 2.0 V 0.15 A1%+2
2% + 1
1.2345 0.322456 1.2 0.4
0.0001