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Rina8888 [55]
3 years ago
13

Which of the following planets orbits the fastest around the sun?

Physics
1 answer:
Snezhnost [94]3 years ago
8 0

Answer: mercury !

Explanation:

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A person running has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?
allsm [11]

Explanation:

Momentum is mass times velocity.

p = mv

720 kg m/s = m (5 m/s)

m = 144 kg

7 0
3 years ago
Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
MariettaO [177]

Answer:

The separation between the two lowest levels =  1.24 * 10^{-39}J

The values of n where the energy of molecule reaches 1/2 kT at 300K = 2.2 * 10^{9}

The separation at this level = 1.8 * 10^{-30}J

Explanation:

Knowing the formula

En = \frac{n^{2} h^{2}  }{8 mL^{2} }

Mass of oxygen molecule

m (O2) = 32 amu * \frac{1.6605 * 10^{-27 kg} }{1 amu}

So the energy diference between the two lowest levels:

E2 - E1 = \frac{3h^{2} }{8mL^{2} }

E2 - E1 =  \frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2}   } = 1.24 * 10^{-39}J

Now we should find n where the energy of molecule reaches 1/2 kT

En = \frac{n^{2} h^{2}  }{8 mL^{2} } = \frac{1}{2}kT

\frac{h^{2}  }{8 mL^{2} } = 4.13 * 10^{-14}J

n^{2} *  (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K

n = 2.2 * 10^{9}

by the end is necessary to calculate the separation of the level

En - En-1 = (n^{2} - (n - 1)^{2}) * \frac{h^{2}  }{8 mL^{2} }

              = 1.8 * 10^{-30}J

4 0
3 years ago
Compare the Summer Solstice with the Autumn Equinox. Justify your response in two or more complete sentences.
ryzh [129]
Midway between the two<span> solstices we have equinoxes – Vernal Equinox in March and </span>Autumnal Equinox<span> in September. ... After this time, the Earth's northern axis is tilted </span>more<span> and </span>more<span>towards ... Then on </span>Summer Solstice<span>, the Sun will reach its farthest north position in the sky</span>
4 0
3 years ago
Read 2 more answers
A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied
Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that  the minimum force that must be applied on the <em>book is</em>

  • F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the  Force  is mathematically given as

F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

For more information on this visit

brainly.com/question/23379286

8 0
2 years ago
A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet
kogti [31]
The centripetal force, Fc, is calculated through the equation, 
                                    Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius. 
Substituting the known values,
                                     Fc = (112 kg)(8.9 m/s)² / (15.5 m)
                                         = 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N. 
4 0
3 years ago
Read 2 more answers
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