If a sound wave speeds up and its frequency does not change, what happens to the wavelength?

1 answer:

5 0

Answer: No, the speed of propagation is constant in a given medium; only the wavelength changes as the frequency changes.

Explanation:

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Which of the following correctly describes one way that photosynthetic plants are involved in the carbon cycle? Choose 1 answer:

brilliants [131]

Answer:

The correct answer is - (Choice A) A Photosynthetic plants can move carbon from the atmosphere into an ecosystem’s food web.

Explanation:

Photosynthesis is the process performed by photosynthetic or green plants in presence of light atmospheric carbon dioxide and water converted into glucose (6 carbon compound) and oxygen.

This glucose or sugar is utilized by the plants that are producers in the ecosystem food web. These producers are fed by consumers and carbon molecules transfer from all the trophic levels in the food chain or food web in the process.

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3 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$