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lisabon 2012 [21]
3 years ago
9

You are hiking in Yellowstone National Park. A park ranger is speaking to students visiting a hot spring. Temperatures there can

reach as high as 456° Fahrenheit. She said some microscopic organisms can live in the hot springs. In what domain do these organisms most likely belong?
A.
Eubacteria

B.
Animal

C.
Archaebacteria

D.
Protist
Physics
1 answer:
Sholpan [36]3 years ago
3 0
I’m pretty sure it’s C Archaebacteria
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Convert 2 days into second​
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Answer:

<em><u>172,000 second </u></em>

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3 years ago
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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2010 × 103 seconds (about
vazorg [7]

Answer:

Radial acceleration of moon is a_{r} = 2.246\times 10^{-3}\frac{m}{s^{2} }

Explanation:

Given :

Time period T = 1.987 \times 10^{6} sec

Distance from center of moon to planet r = 225 \times 10^{6} m

From the equation of radial acceleration,

  a_{r} = r\omega ^{2}

Where \omega = 2\pi f = \frac{2\pi }{T}

So   \omega = 3.16 \times 10^{-6} \frac{rad}{s}

Now moon's radial acceleration,

 a_{r} = 225 \times 10^{6} \times (3.16 \times 10^{-6} )^{2}

 a_{r} = 2246.76 \times 10^{-6}

 a_{r} = 2.246\times 10^{-3} \frac{m}{s^{2} }

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3 years ago
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The higher the pressure, the higher boiling point of water. At lower the pressure, the boiling point of water comes down. So, the lower pressure inreases the boiling resulting more evaporation. As we go higher in altitude, the atmospheric pressure decreases. This results in decreasing the boiling point at higher altitude and increase in boiling of water. In fact, at the sea level ,the the sea water boils at 100 degree C where atmospheric pressre is normal. However , the boiling takes place at a lower temperature at the top of a mountain due to low pressure. In other words the boling is faster at the top of a mountain than that at its foot.
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Explanation:

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(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz

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