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3 years ago

10

What is a likely consequence of preventing prescribed burns to forest ecosystems?

1 answer:

liraira [26]3 years ago

5 0

The forest ecosystem will regain back its natural ecology compared to burning of the forest and other illegal activities done to the forest. And also, some of the animals living in that forest will live and those that are endangered will not be threatened.

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Newton first law of motion ?

maw [93]

Answer:

The law of inertia

Explanation:

A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force

8 0

3 years ago

Read 2 more answers

Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus

ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

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8 0

2 years ago

Atoms with a low ionization energy hold tight to their outer valence electrons.

Anestetic [448]

True would be the Correct answer

5 0

4 years ago

Read 2 more answers

A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i

Darya [45]

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity $v$ , is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

$\implies v=\sqrt{hg}$

8 0

4 years ago

A cyclist is traveling 10 m/s with an acceleration of 6 m/s2. How fast is the cyclist traveling at the end of 20 seconds

krok68 [10]

Answer:

<em>The cyclist is traveling at 130 m/s</em>

Explanation:

<u>Constant Acceleration Motion </u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

$v_f=v_o+at$

The cyclist initially travels at 10 /s and it's accelerating at a=6m/s^2. We need to know the new speed when t= 20 seconds have passed.

Apply the above equation:

$v_f=10+6\cdot 20$

$v_f=10+120$

$v_f=130\ m/s$

The cyclist is traveling at 130 m/s

7 0

3 years ago