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madreJ [45]
3 years ago
8

Why does the fact that Americans are using less fertilizer not indicate a decline in American farming?

Chemistry
2 answers:
meriva3 years ago
4 0
It shows that the population is moving towards a more organic and natural way of growing crops. Rather than using fertilizer and putting toxins on the plants that we eat from, we are getting smarter and growing food naturally. Thus not killing ourselves in the process.

Svetllana [295]3 years ago
4 0

Answer:

The decrease in fertilizer use does not mean that there is a decrease in agricultural production in America. On the contrary, there could be an increase in production even using fewer fertilizers.

The reason is simple, THE TECHNOLOGY.

Explanation:

Currently there are many new methods to save fertilizer products in the field and make crops more efficient. This is called precision agriculture.

With these methods, for example, images taken with drones or satellites (with multispectral or hyperspectral cameras) can be used to assess the status of a crop, the lack of nutrients from the soil, and even if there are pests or diseases on the plats; All this allows the farmer to apply the fertilizer or other products, only in the area that presents the problem and not in the entire crop.

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. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r
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Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

Explanation : Given,

Wavelength = 435.8nm=435.8\times 10^{-9}m

conversion used : 1nm=10^{-9}m

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency

h = Planck's constant = 6.626\times 10^{-34}Js

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c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}

E=4.56\times 10^{-19}J

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

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