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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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<h2>Answer: 12 s</h2>

Explanation:

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y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

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y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

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As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

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