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alina1380 [7]
4 years ago
11

The Great Lakes region has both biotic and abiotic components. Abiotic components include water, temperature, rainfall, and geol

ogical features. Biotic components include all of the animal and plant species in the area. The ecosystem of the Great Lakes includes _______. A. only the abiotic components of the Great Lakes region B. only the animal species that live in the Great Lakes region C. all of the biotic and abiotic components of the Great Lakes region D. only the plant and animal species that live in the Great Lakes region
Physics
2 answers:
Andrew [12]4 years ago
3 0

the answer is C. all of the biotic and abiotic components of the great lakes region

Rudiy274 years ago
3 0

Answer:

C. all of the biotic and abiotic components of the Great Lakes region

Explanation:

because i just did it no study island

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From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 121.6 nanome
lina2011 [118]

Answer:

b) The star is moving away from us.

Explanation:

If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.

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3 years ago
How does force and acceleration work together to create motion?
marta [7]
When a force acts (pushes or pulls) on an object, it changes the object's speed or direction (in other words, makes it accelerate). The bigger the force, the more the object accelerates. When a force acts on an object, there's an equal force (called a reaction) acting in the opposite direction.
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Ligaments are important because they provide joints with:
lora16 [44]

Answer:

the answer is B. it's too easy

4 0
4 years ago
1) At which lettered point or points is the object moving the fastest?
emmainna [20.7K]

Here we have a time-position graph with some points marked on it, and we want to analyze some points in it.

1) The object is moving the fastest in the point where the slope of the curve is the largest.

In this case, the point is D.

2) Remember that the graph evolves to the right, so, if at any point, the right side is above the left side, then at that particular point the object is moving to the left (negative x-axis)

In this case, the points are:

C, D, and E.

3) The object speeds up when the curvature of the curve changes and the slope increases, for example, from B onwards, the speed increases.

Then at point C the speed is increasing.

4) Opposite to what we said before, if the slope decreases then the object is slowing down.

In this case we can see that the slope decreases after A and after E.

5) When the object is turning around?

This is easy, we can see that the object goes to the right (positive x-axis) until it reaches B and then goes back, then at point B the object turns around.

If you want to learn more, you can read:

brainly.com/question/11290689

8 0
3 years ago
Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
8 0
3 years ago
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