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Inga [223]
3 years ago
6

The diagram shows a vertical number line. what is the position of point X?

Physics
2 answers:
mihalych1998 [28]3 years ago
8 0

The correct answer is option 4, 16.

Explanation

Each division on the number line represents 4 units. The point X is located 4 divisions above 0, the origin. On this particular number line, 4 divisions represents 4\times 4=16 units. This means that the point X is located at y=16 on the number line.

The correct answer is the 4th option, whose value is 16.

postnew [5]3 years ago
5 0

As per the question the diagram given is a vertical line.

We are asked to calculate the position of X.

Here the vertical line is divided into many segments.

As per the diagram, it is obvious that the points marked at different positions are separated by four units from each other i.e the distance between two successive marks is 4 units.

The point X is situated at four marks away from 0.

Hence the position of X will be= 4×4 units

                                                    = 16 units

Hence the correct answer of this question is D.

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Scientists might make a computer model of volcanic eruptions. What is the
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The Jensens dccided to spend their family vaeation white water rafting. During one segment of their trip down the river, the raf
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24,187.04 J ≈ 24,200 J

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final speed (v) = 15.2 m/s

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3 years ago
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Answer:

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Explanation:

Please give me brainliest :) I will greatly appreciate it :D

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2 years ago
Read 2 more answers
What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
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Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

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\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

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\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

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