1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Margarita [4]
2 years ago
10

Mountain Ecosystems Assignment Answer each question below in complete sentences.1) Identify a predator from an Ethiopian Highlan

ds ecosystem. How else can this animal be classified in terms of feeding relationships? What is its prey? How else can its prey animal be classified in terms of feeding relationships? Identify a predator from a Himalayan Mountain ecosystem. How else can this animal be classified in terms of feeding relationships? What is its prey? How else can its prey animal be classified in terms of feeding relationships?
Physics
1 answer:
Marta_Voda [28]2 years ago
8 0

An example of a a predator in the Ethiopian Highlands ecosystem is the

Wolf.The wolf is a carnivore which acts as a predator to other smaller

animals in the ecosystem.

<h3>What is a Prey?</h3>

Preys are mostly smaller animals in which the predators feed on for food.

They are usually herbivores and primary consumers in the ecosystem.

Examples of Preys include:

  • Rodents
  • Deer
  • Bison etc.

Read more about Feeding relationship here brainly.com/question/9852437

You might be interested in
In a light wave, what properties tell you the color of light
sesenic [268]

Answer:

Answer :- In a light wave the property of wave which tells about the color of light is it's Wavelength .

Wavelength is the distance between one crest and one through , also it is the distance after which the wave repeat itself !

It's SI unit is meter !

It is scalar quantity !!

Different Wavelength of light have different color !!

• VIBGYOR

i.e, Violent , Indigo , Blue , Green , Yellow Orange, and Red along with their shades are the colors which we can see !!

• They almost range from 400nm to 700nm ( visible range of light ) !!

4 0
3 years ago
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
2 years ago
The diagram shows a positive charge moving in a magnetic field.
slavikrds [6]

By the right hand rule the magnetic force on the charge acts up

6 0
3 years ago
Read 2 more answers
Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is
suter [353]

Answer: 0.223 m/s^{2}

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity g of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

F=G\frac{m_{E}m}{r^2}  (1)

Where:

F is the module of the force exerted between both bodies

G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}} is the gravitational constant

m_{E}=5.98(10)^{24} kg is the mass of the Earth

m are the mass of each communications satellite

r=R_{E}+h is the distance between the center of the Earth and the satellite

R_{E}=6.38(10)^{6} m is the radius of the Earth

h=3.59(10)^{7} m is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

F=mg  (2)

Combining (1) and (2):

G\frac{m_{E}m}{r^2}=mg  (3)

Isolating g:

g=\frac{G M_{E}}{r^2}  (4)

Remembering r=R_{E}+h:

g=\frac{G M_{E}}{(R_{E}+h)^2}  (5)

g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}  

Finally:

g=0.223 m/s^{2}  

5 0
3 years ago
How does the density of most metals compare to most non metals?
Neko [114]
Harder. Not compressible(unless using an extremely strong force). Non-metal have more of a chance of breaking than metals.
8 0
3 years ago
Other questions:
  • Jody has a mass of 50 kilograms what is his weight on earth
    14·2 answers
  • Which of the following represents the wavelength of a wave?<br> A <br> B <br> C <br> D
    10·1 answer
  • What are TWO benefits of scientific using a diagram to model the water cycle? [PLS HELP ASAP]~pick 2 letters~
    9·1 answer
  • How do sound waves travel through a medium?
    15·1 answer
  • Learning Goal: To understand how to find the wavelength and diffraction patterns of electrons.
    12·1 answer
  • A water tank has a square base of area 8 square meters. Initially the tank contains 70 cubic meters. Water leaves the tank, star
    14·1 answer
  • What is responsible for creating distinct weather patterns?
    11·1 answer
  • Think about a situation at school where you or someone you know
    10·1 answer
  • Why do you think the Founders divided the process into two steps (impeachment and conviction)?
    7·2 answers
  • Which change of state is shown in the photo
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!