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2 years ago

A swarm of locust eats the leaves of most of the trees in a forest. How would the trees be most directly affected?

Physics

1 answer:

Sphinxa [80]2 years ago

5 0

Answer:

A trees food and air source would be deprived because of this. Trees need leaves to perform photosynthesis, so besides their roots soaking up water and nutrients, they wouldn't have any other form of food until those leaves grow back. If they don't and the locusts continue to eat up the tree, it will eventually starve and die.

Explanation:

Imagine the trees trunk and roots as your body and the leaves as your mouth/food source. God Forbid you catch a cold or some sickness. You probably won't have the appetite to eat. Despite that, you'll still have to. But, if you get so sick to the point that eating just makes you throw it up before your body can take in the nutrients and beneficial stuff, you will eventually get to the point where eating is nearly impossible. This will lead up to the sickness taking over your body, as there's no energy going in to help it fight it off, and you would die.

(Gah, sorry for the dark theme... didn't think it would turn like that lol)

Anyways, hope this helped!

Source(s) used: N/A

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

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2 years ago

Is an atom with one valence electron more reactive than an atom with two electrone?

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Answer:

An atom with a closed shell of valence electrons (corresponding to an electron configuration s2p6) tends to be chemically inert. An atom with one or two valence electrons more than a closed shell is highly reactive, because the extra valence electrons are easily removed to form a positive ion.Explanation:

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3 years ago

A player kicks a soccer ball from ground level and sons at flying at an angle of 30° at a speed of 26MS how far did the ball tra

RSB [31]

Answer:

Explanation:

60 meters is he answer for this question

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3 years ago

A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a

RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2$

$F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N$

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

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3 years ago

Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests o

Andrew [12]

Answer:

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Explanation:

The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.

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