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Stella [2.4K]
2 years ago
14

Average human body temperature is 98.6°F. What is the equivalent in degrees Celsius?

Physics
1 answer:
Gala2k [10]2 years ago
6 0

37° Celsius is equal to 98.6° Fahrenheit

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A water wave has a speed of 22.0 meters/second if the wave frequency is 0.0680 hertz what is the wavelength
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V = f(wavelength)
22.0 = 0.0680 (wavelength)
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. What will happen if there is an oil spill in the Arctic Ocean based on the statements above? A) The Atlantic and Southern Ocea
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Roxy is driving down East Capitol Street. She drives 1380 meters in 45 seconds. Assuming she does not speedup or slow down, what
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30.66667

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William wanted to create a report on a geographical location with the greatest species diversity. Which ecosystem can consider f
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The ecosystem that he should consider is the tropical rain forest ecosystem.

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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
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(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
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