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3 years ago

Which type of species would be the perfect option to repopulate an area that is in the primary succession stages?

1 answer:

4 0

Answer:In primary succession, newly exposed or newly formed rock is colonized by living things for the first time. In secondary succession, an area previously occupied by living things is disturbed—disrupted—then recolonized following the disturbance.The first organisms to appear in areas of primary succession are often mosses or lichens. These organisms are known as pioneer species because they are the first species present; pioneer species must be hardy and strong, just like human pioneers.A heterotroph is an organism that eats other plants or animals for energy and nutrients. The term stems from the Greek words hetero for “other” and trophe for “nourishment.” Organisms are characterized into two broad categories based upon how they obtain their energy and nutrients: autotrophs and heterotrophs.

Explanation:I forgot Extinct

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A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

5 0

4 years ago

Why is random sampling and random assignment in experimental research important?

Hunter-Best [27]

So the result is not biased or affected in some way

6 0

3 years ago

Please answer :><br> 40 POINTS

LenaWriter [7]

Answer:

rotates

Explanation:

I'm so bored

yrfgggghhgghhyuj

8 0

3 years ago

Read 2 more answers

A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c

mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0

3 years ago

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h

Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2$

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft$

The minimum stopping distance is 141.78 ft.

4 0

3 years ago