All categories

3 years ago

The phosphorus cycle is important to ecosystems. Choose all of these statements that are true concerning the phosphorus cycle.

Physics

2 answers:

Lesechka [4]3 years ago

7 0

Answer:

The largest reservoir of phosphorous is sedimentary rock.

Major sources of phosphorous to aquatic ecosystems are fertilizer runoff, sewage leaks, and industrial wastes.

Eccess phosphorous can lead to eutrophication

Explanation:

Phosphorus come from different sources such as aquatic ecosystems and fertilizers used for plants. When these substances containing phosphorus and those from industrial wastes find their way into water bodies, they tend to cause eutrophication, which is the natural enrichment of water bodies.

Also, it is known that a very small portion of phosphoric acid contribute to acid rain in the atmosphere.

Helga [31]3 years ago

7 0

Answer:

The statements that are true are:

-The largest reservoir of phosphorous is sedimentary rock

- Major sources of phosphorous to aquatic ecosystems are fertilizer runoff, sewage leaks, and industrial wastes.

- Excess phosphorous can lead to eutrophication.

Explanation:

In nature, phosphorous is found in form of phosphate ions PO₄⁻³, which are forming part of sedimentary rocks. Upon the action of wind and rain, phosphorous is washed into the soil (it dissolves and its passes into the soil in form of phosphate compounds). So, the largest reservoir of phosphorous is sedimentary rock.

Phosporous in the soil is absorbed by animal and plants and it is transformed in biological components (like nucleic acids, phospholipids, etc), and when animal and plants die, phosphorous return to the soil. So, the most of the phosphorous cycle occurs in the Earth (soil, water, living beings). In fact, phosphorous does not circulate through atmosphere.

Human activities such as farming and industry impact in phosphorous cycle. Fertilizers, wastes from human activities and phosphorous-containing products are poured in oceans and they feed the cycle.

But the excess of phosphorous often leads to an overgrowth of algae, which causes a lack of oxygen in water and consequenty the death of aquatic organisms. This fenomena is called eutrophication.

Finally, the major acid found in acid rain is sulphuric acid (it is not phosphoric acid).

You might be interested in

Three cannons are located at the top of a cliff above a level plain. Cannon A is aimed at an angle of 25° above the horizontal a

vodka [1.7K]

Answer:

The canon B hits the ground fast.

Explanation:

Given that,

Speed of cannon A = 85 m/s

Speed of cannon B= 100 m/s

Speed of cannon C = 75 m/s

We need to calculate the cannonballs will hit the ground with the greatest speed

Using conservation of energy

The final kinetic energy of canon depends on initial kinetic energy and potential energy.

The final velocity depends upon initial velocity and initial height.

So, the initial velocity of canon B is high.

Hence, The canon B hits the ground fast.

3 0

3 years ago

When a wave strikes a solid barrier it behaves like a basketball hitting a backboard this wave behavior years called?

Mama L [17]

This behavior is called reflection.
Reflection is a change of in direction of the wave when it reaches another medium. Imagine a wave colliding with a glass in a tank of water.
During reflection, some of the initial energy of the wave is lost.
Waves always reflect with at same angle at which it approached the obstacle.

5 0

3 years ago

Use the crisscross method to find the chemical formula for the ionic compound formed by strontium (Sr) and bromine (Br).

nignag [31]

Answer:it’s c SrBr2

Explanation:

8 0

3 years ago

Read 2 more answers

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Alex787 [66]

Answer:

a) 1.092 m/s

b) 0.33 m

c) 0.25 m

Explanation:

To start with, from the formula of wave, we know that

v = f λ, where

v = velocity of wave

f = frequency of the wave

λ = wavelength of the wave

Again, on another hand, we know that

T = 1/f, where T = period of the wave

From the question, we are given that

t = 2.7 s

d = 0.66 m

λ = 5.9 m

Period, T = 2 * t

Period, T = 2 * 2.7

Period, T = 5.4 s

If T = 1/f, then f = 1/T, thus

Frequency, f = 1/5.4

Frequency, f = 0.185 hz

Remember, v = f λ

v = 0.185 * 5.9

v = 1.092 m/s

Amplitude, A = d/2

Amplitude, A = 0.66/2

Amplitude, A = 0.33 m

If the other distance travelled by the boat is 0.5, then Amplitude is

A = 0.5/2

A = 0.25 m

6 0

3 years ago

What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?

VikaD [51]

(a) $1.33\cdot 10^{-11} m$

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

$e \Delta V = \frac{hc}{\lambda}$

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

$e=1.6\cdot 10^{-19}C$ is the electron's charge

$\Delta V = 93.3 kV = 93300 V$ is the potential difference

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.00\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the emitted photon

Solving the formula for $\lambda$ , we find:

$\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m$

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

$E=\frac{hc}{\lambda}$

where

h is Planck constant

c is the speed of light

$\lambda=1.33\cdot 10^{-11} m$ is the wavelength of the photon, calculated previously

Substituting,

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J$

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

$E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV$

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

$e \Delta V = \frac{hc}{\lambda}$

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0

3 years ago