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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 dog, that has a mass of 16 kgs, runs across a yard. What is the average force applied to the dog from the ground as it runs ac

VikaD [51]

Answer:

156.96 N

Explanation:

F=ma where m is the mass and a is acceleration

Substituting 16 Kg for m and 9.81 m/s2 for g then

F=16*9.81= 156.96 N

4 0

3 years ago

In some areas, farmers have installed windmills to generate electricity to run the farm.

Aleks [24]

Answer:

Number 3

Explanation:

Unlike other numbers, this states that wind is "renewable". The choice 2 sounds a bit selfish, or not worded properly, I believe choice 3 is the answer. However, you are the judge of choosing to believe this or not.

Good luck!

8 0

2 years ago

A sled is moving at a constant speed down a surface inclined at 45.0° with the horizontal and travels 30 meters in 4 seconds. Ca

slamgirl [31]

To obtain the vertical component of the velocity, it is necessary to multiply the velocity at this time by sin 45 degrees. Since the velocity is not given, it must be solved first by dividing the distance by the time. To show this:

Velocity = distance / time
Velocity = 30 m / 4s
Velocity = 7.5 m/s

Vertical velocity = 7.5 sin 45
Vertical velocity = 5.3 m/s

5 0

3 years ago

Why do I need to use the Unit Circle in Physics? And how do I use it?

stiks02 [169]

Answer:

The unit circle helps in making so many calculations and equations easy.

Explanation:

You cannot separate the knowledge of trigonometry to application in equations to physics. The unit circle is known to have a radius of one. This means that the distance from the centre of the circle, regardless of the unit of measurement, to any point of the edge of the circle is 1. Since the unit circle is very helpful in trigonometry, and trigonometry in turn is the projection of triangles and angles that is very crucial in the calculation of momentum, velocity and other factors of physics, the importance of the unit circle cannot be overemphasized in physics.

6 0

2 years ago

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in t

Liono4ka [1.6K]

(a) 5.69 N/C, vertically downward

We can calculate the acceleration of the electron by using the SUVAT equation:

$d=ut+\frac{1}{2}at^2$

where

d = 4.50 m is the distance travelled by the electron

u = 0 is the initial velocity of the electron

$t=3.00 \mu s = 3.0 \cdot 10^{-6} s$ is the time of travelling

a is the acceleration

Solving for a,

$a=\frac{2d}{t^2}=\frac{2(4.50)}{(3.0\cdot 10^{-6})^2}=1.0\cdot 10^{12} m/s^2$

Given the mass of the electron,

$m=9.11\cdot 10^{-31} kg$

We can find the electric force acting on the electron:

$F=ma=(9.11\cdot 10^{-31})(1.0\cdot 10^{12})=9.11\cdot 10^{-19}N$

And the electric force can be written as

$F=qE$

where

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

E is the magnitude of the electric field

Solving for E,

$E=\frac{F}{q}=\frac{9.11\cdot 10^{-19}}{-1.6\cdot 10^{-19}}=-5.69 N/C$

The negative sign means that the direction of the electric field is opposite to the direction of the force (because the charge is negative): since the force has same direction of the acceleration (vertically upward), the electric field must point vertically downward.

(b) Yes

We can answer the question by calculating the magnitude of the gravitational force acting on the electron, to check if it is relevant or not. The gravitational force on the electron is:

$F=mg$

where

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.81 m/s^2$ is the acceleration due to gravity

Substituting,

$F=(9.11\cdot 10^{-31})(9.81)=8.93\cdot 10^{-30}N$

We see that the gravitational force is basically negligible compared to the electric force calculated in part (a), therefore we can say it is justified to ignore the effect of gravity in the problem.

7 0

3 years ago