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ser-zykov [4K]
3 years ago
15

How do you think the formation of new stars is related

Physics
1 answer:
Rufina [12.5K]3 years ago
4 0

Answer:

when the matter from old stars reform

Explanation:

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Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p
DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = \frac{mv^{2}}{r}

Electrostatic force = \frac{kq^{2}}{r^{2}}

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}

v=\sqrt{\frac{kq^{2}}{mr}}

v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by  2.068 x 10^6 m / s.

6 0
3 years ago
Which best describes a difference between laser light and regular light?
Inga [223]
The basic difference is that the ordinary sources are incoherent that means that the discrete frequencies merge up to give an intermediate between the maximum and minimum frequencies. While the laser is coherent containing the single frequency with maximum amplitude. thus travelling far.
4 0
3 years ago
If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh
Gala2k [10]

the correct answer is c

4 0
3 years ago
A golf ball is dropped from rest from a height of 9.5m.  It hits the pavement then bounces back up rising  just 5.7 m before fal
Oksanka [162]

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

4 0
3 years ago
Read 2 more answers
Calculate the acceleration of a student riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds.
solong [7]
Acceleration is equal to velocity final minus velocity initial divided by time. 6m/s minus 4m/s divided by 5 seconds is 0.4m/s^2.

7 0
3 years ago
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