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Pani-rosa [81]
3 years ago
15

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]3 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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The temperature of a gas is increased by 10degree Celsius. what is the corresponding change on Fahrenheit scale?​
Grace [21]

On a Fahrenheit thermometer, the gas becomes 18 degrees warmer.

8 0
2 years ago
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the
san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
Please help!?!?!?!?!?!?!
adoni [48]

Answer:

D

Explanation:i think but dont get mad if im wrong

5 0
2 years ago
In need help I need someone that is really good at physics
Aloiza [94]
10/70×360°
=51.4°

hope thus helps
5 0
3 years ago
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