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2 years ago

15

How many non square lie be 2² and 3²?

1 answer:

erastovalidia [21]2 years ago

7 0

Answer:

on

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How many non-square numbers lie between the squares of 12 and 13?

Answer

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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.

Complete step-by-step answer:

In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.

Square of 12 = 122=144 and square of 12 = 132=169

As 12 and 13 are consecutive numbers so all numbers between their squares will be non-square numbers.

Therefore, 169 – 144 = 25

Total number of numbers between 169 and 144 (i.e., excluding 144 and 169) = 25 – 1 = 24.

Explanation:

Brian least po please

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if the radius of the capillary tube is doubled, what changes will take place in the hieght of rise of liquid with capacity tube

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2 years ago

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,

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

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m$

So,

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

Also, $\Delta E=h\times \nu$

So,

$h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}$

$\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

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

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3 years ago

An object begins x=75.2 m and undergoes a displacement of -48.7 m. what is its final position?

xz_007 [3.2K]

Answer:

26.5 m

Explanation:

$x_{o}$ = initial position of the object = 75.2 m

$x$ = final position of the object

$d$ = displacement of the object = - 48.7

Displacement of the object is given as the difference of final and initial position of the object

$d = x - x_{o}$

Inserting the values

- 48.7 = x - 75.2

x = 26.5 m

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