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Sauron [17]
3 years ago
10

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

centripetal acceleration of 9.0 x 1022 m/s2.
a) Find its speed.
b) In an excited state with principle quantum number n = 10, the electron is 100 times farther away, and its centripetal acceleration is 10,000 times.
Physics
1 answer:
Readme [11.4K]3 years ago
8 0

Answer:

Explanation:

Given

radius of electron(r)=5.3\times 10^{-11} m

centripetal acceleration (a_c)=9\times 10^22 m/s^2

we know

a_c=\frac{v^2}{r}

v=\sqrt{r\times a_c}

v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}

v=\sqrt{47.7\times 10^11}

v=21.84\times 10^5 m/s

(b)For n=10

r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m

a_c=10^4\times 9\times 10^{22} m/s^2

a_c=9\times 10^{26} m/s^2

v=\sqrt{r\times a_c}

v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}

v=21.84\times 10^8 m/s

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Answer:

A

Explanation:

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Find the time it takes for each object to accelerate from 0m/s to 40 m/s when pushed with 100N of force
Ivan

Answer:

40s

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A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
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Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

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Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

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Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

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Therefore, it will take the kangaroo 1.08 s to return to the earth.

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