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monitta
4 years ago
6

What's your normal body temperature? it may not be 98.6°f, the oft-quoted average that was determined in the nineteenth century.

a more recent study has reported an average temperature of 98.2°f. what is the difference between these averages, expressed in celsius degrees?
Physics
3 answers:
harkovskaia [24]4 years ago
6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

Igoryamba4 years ago
6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

givi [52]4 years ago
6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC OR

98.6*5ºC/9ºF = 54.80 98.25ºC/9ºF = 54.60 54.80 - 54.60 = .2ºC

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3 years ago
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Answer:

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I = \frac{m*v}{e*\mu_{0}*n*r} = \frac{9.1 \cdot 10^{-31} kg*3.0 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*4\pi \cdot 10^{-7} T.m/A*3500/m*0.03 m} = 0.0129 A  

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I hope it helps you!

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