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?I think we want to say:x
Delta R/R = 3.9E-3*DeltaT
=3.9E-3*50 = .2 = 20%
-----Original Message-----
From: brian whatcott
Sent: Saturday, February 05, 2011 4:53 PM
To: phys-l@carnot.physics.buffalo.edu
Subject: Re: [Phys-l] real-world problem and other questions about Car Talk
Nope. That's not right either! :-)
Brian W
On 2/5/2011 3:46 PM, brian whatcott wrote:
3.9E-3 DeltaR/DegC X 5E1 DegC = 2.0E-2 deltaR??_______________________________________________
Brian W
On 2/5/2011 9:49 AM, Bernard Cleyet wrote:
I suspected not enuf change in resistance, but calc'd and find quite_______________________________________________
reasonable. The starter on oldies is bolted to a frame, as part of the
engine case, so reasonable conductive path from the cylinders. So a 50 C
change is not unreasonable, which results in an ~ 20% change in Cu
resistivity.
[3.9 E-3 /C]
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