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ivanzaharov [21]
3 years ago
9

Which event occurred approximately 4.6 billion years ago

Physics
1 answer:
Levart [38]3 years ago
4 0
The answer to this is formation of earth and our solar system
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In this example we see how thermal expansion can actually throw off the accuracy of a length-measuring device—namely, a tape mea
Ulleksa [173]

Answer:

The new length is 50.00885m

Explanation:

linear thermal expansion coefficient Fe 11.8e-6 /K

The new length can be determined using the following equation:

∆L/L = α∆T, where α is linear thermal expansion coefficient

∆L = Lα∆T = 50(11.8e-6)(35-20) = 0.00885 m

New length = ( 50.000 + 0.00885)m =

New length = 50.00885 m

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Answer:

Period = 1.33 seconds

Explanation:

Period = 1/0.75

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3 years ago
The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
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Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

v_{rms} = \sqrt{ \frac{3RT}{M}}

So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

v_{rms,o2}^2 *M_{o2} =3RT_{o2}

solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}

T_{o2} =1.58*10^5 K

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Where is the “part 2” to view ?
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A sign is supported by two ropes of different length attached to each of the two upper corners. One rope has a force of 36.28 N
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