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marta [7]
3 years ago
12

Select the correct answer. x y 2.5 6.25 9.4 88.36 15.6 243.63 19.5 380.25 25.8 665.64 The table lists the values for two paramet

ers, x and y, of an experiment. What is the approximate value of y for x = 4? A. 11 B. 16 C. 24 D. 43
Physics
1 answer:
In-s [12.5K]3 years ago
6 0

Answer: B. 16

Explanation:

The given table:

x y

2.5 6.25

9.4 88.36

15.6 243.63

19.5 380.25

25.8 665.64

Here, we observed that value of y is the square value of x.

i.e. y=x^2  [For example 2.5²=6.25, 9.4²=88.36, 15.6²=243.63]

Put x=4, we get

y=4^2=16

Hence, the approximate value of y for x = 4 is 16.

So, the correct option is B. 16 .

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Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

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The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

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e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

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Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

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r=5.3\cdot 10^{-11}m

while the electron mass is

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and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

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