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harina [27]
3 years ago
12

The graph of a line gets ________ as the value of the slope gets smaller.

Physics
2 answers:
Sav [38]3 years ago
8 0

Answer: Option c: <u>Less steep</u>

Explanation:

The graph of a line gets less steep, the value of slope gets smaller.

The slope describes the steepness in the graph. Larger slope means steeper graph line. On the other hand, smaller graph line means that the slope is less steep. Hence, the correct answer is The graph of a line gets <u>Less steep</u> as the value of the slope gets smaller.

klemol [59]3 years ago
4 0

The graph of a line gets less steep as the value of the slope gets smaller.

 

<span>A </span>line<span> is a straight one-dimensional figure having no thickness and extending infinitely in both directions. A </span>line<span> is sometimes called a straight </span>line<span> or, more archaically, a right </span>line<span> (Casey 1893), to emphasize that it has no "wiggles" anywhere along its length.</span>

 

The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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While riding on I-26 towards Columbia, your mother slams on the brakes, to avoid hitting the car in front of her. Both of your b
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A mis-hit golf ball flies straight upward. It reaches a height of 31.0 meters before falling back down. How fast was it going wh
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now we can use kinematics to find the final speed of the ball when it will hit back the ground

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5 0
3 years ago
10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.
Xelga [282]

Answer:

\lambda=1282\ nm

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

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

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

So,  

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

Given, n_i=5\ and\ n_f=3

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m

\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m

\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m

\lambda= 1.2828\times10^{-6}

1 m = 10⁻⁹ nm

\lambda=1282\ nm

6 0
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