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3 years ago

If transpiration didn't take place, would water be able to enter the roots of the plant? Explain your reasoning

Physics

1 answer:

s2008m [1.1K]3 years ago

3 0

Answer:

If transpiration didn't take place water would still be able to enter the roots of a plant

Explanation:

transpiration is the process of water leaving from living organisms to the atmosphere, therefore, if transpiration didn't occur the water would not transpire to the atmosphere and would remain in the root but water absorption would not change because it is a biological need for the living organism as such

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What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?

PolarNik [594]

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,

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

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

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

8 0

3 years ago

True or false. Atoms of a given element will have the same mass.

jarptica [38.1K]

Answer:

True

Explanation:

Atoms of the same element will always have the same atomic mass or the mass listed in the periodic table.

4 0

3 years ago

Which one of the following devices converts radioactive emissions to light for detection?

Finger [1]

the radiotracer converts radioactive emissions to light for detection. the answer is D.

6 0

3 years ago

Rolled cookies are cut into fancy shapes.<br><br> True<br> False

Paha777 [63]

Answer:

false

Explanation:

the answer is false

5 0

2 years ago

Read 2 more answers

Define wind ward side

Rasek [7]

Answer:

The windward side is that side which faces the prevailing wind (upwind), whereas the leeward, or "lee" side, is the side sheltered from the wind by the mountain's very elevation (downwind)

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3 years ago