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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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Is the magnitude of the acceleration of the center of mass of spool A greater than, less than, or equal to the magnitude of the
Anon25 [30]

Answer:

The acceleration of the centre of mass of spool A is equal to the magnitude of the acceleration of the centre of mass of spool B.

Explanation:

From the image attached, the description from the complete question shows that the two spools are of equal masses (same weight due to same acceleration due to gravity), have the same inextensible wire with negligible mass is attached to both of them over a frictionless pulley; meaning that the tension in the wire is the same on both ends.

And for the acceleration of both spools, we mention the net force.

The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.

For this system, the net force on either spool is exactly the same in magnitude because the net force is a difference between the only two forces acting on the spools; the tension in the wire and their similar respective weights.

With the net force and mass, for each spool equal, from

ΣF = ma, we get that a = ΣF/m

Meaning that the acceleration of the identical spools is equal also.

Hope this Helps!

6 0
3 years ago
Optimus Prime is flying straight up at 24 m/s when he accidentally drops his mega-ray blaster and it falls 94 m to the ground be
natali 33 [55]

Answer:

The time it will take the mega-ray blaster to hit the ground is 2.57 s.

Explanation:

Given;

initial velocity of Optimus Prime, u = 24 m/s

height of fall of the mega-ray blaster, h = 94 m

The time of fall of the mega-ray blaster is calculated using the following kinematic equation;

h = ut + \frac{1}{2}gt^2\\\\94 = 24t +  \frac{1}{2}(9.8)t^2\\\\94 = 24t + 4.9t^2\\\\4.9t^2 +24t -94 = 0\\\\Use \ formula \ method \ to \ solve \ for \ "t"\\\\a = 4.9 , b = 24, c = -94\\\\t = \frac{-b \ +/- \ \sqrt{b^2 -4ac} }{2a} \\\\t =  \frac{-24 \ +/- \ \sqrt{(24)^2 -4(-94 \times4.9)} }{2(4.9)} \\\\t = \frac{-24 \ +/- \ \sqrt{2418.4} }{9.8}\\\\t = \frac{-24 \ +/- \ 49.177 }{9.8}\\\\t = \frac{-24 \ +\  49.177 }{9.8} \ \ or \ \ t = \frac{-24 \ -\  49.177 }{9.8} \\\\

t = 2.57 \ s \ \ or \ \ t = -7.47 \ s

t = 2.57 s

Therefore, the time it will take the mega-ray blaster to hit the ground is 2.57 s.

3 0
3 years ago
In order to increase the efficiency of the detection-to-engagement sequence, one can never have too much information from sensor
Elan Coil [88]

Answer:

False

Explanation:

Much information can be gotten from sensor due to the much data being gathered.

7 0
3 years ago
A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080
deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}

now we have

M = 0.120 kg

m_1 = 0.02 kg

m_3 = 0.08 kg

now we have

I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}

so we have

I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162

I = 0.02835

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}

\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)

\omega = 4.32 rad/s

Now speed of 0.08 kg mass when it reaches to bottom point is given as

v = \omega \frac{L}{2}

v = 4.32 (0.45)

v = 1.94 m/s

3 0
4 years ago
A 30 ky child running at 7 m/s jumps onto a 10 kg sled which was initially at rest. What will be the
Ede4ka [16]
Hope this helps. If you need clarification just ask me!

7 0
3 years ago
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