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Colt1911 [192]
2 years ago
9

Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha

t is the magnitude of the centripetal acceleration in m/sec2? (radius of the circle LaTeX: 5\times 10^{-11}5 × 10 − 11m; period of the motion LaTeX: 1.5 \times 10^{-16}1.5 × 10 − 16sec) Group of answer choices
Physics
1 answer:
Ksivusya [100]2 years ago
8 0

Answer:

A_c=87.73*10^{21}m/s

Explanation:

From the question we are told that

r=5\times 10^{-11}

T=1.5 \times 10^{-16}

Generally the equation for velocity is mathematically given as

Velocity (v)=\frac{2 \pi r}{t}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

Generally the equation for Centripetal acceleration is mathematically given as

A_c=\frac{V^2}{r}

A_c=(\frac{20.944*10^5)}{r5*10^{-11}}

A_c=87.73*10^{21}m/s

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Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

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6 0
3 years ago
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Answer:

t = 5 hr

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