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3 years ago

What is an example of the theory that the best adapted organisms for an environment are the ones most likely to survive

1 answer:

6 0

This is called the theory of evolution.

The theory of evolution is one of the most prominent theories in all of science as it is very far-reaching and all-encompassing. Furthermore, it talks exactly about that. A great example would be that we currently have only different types of organisms in the sea that can breathe under water - they are the best adapted type of organisms for that environment and can therefore prosper in it.

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A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?

garri49 [273]

Answer:

2 m/s^2

Explanation:

a = v^2/r

a = (10m/s)^2 / 50m

a = 2 m/s^2

Leave a like and mark brainliest if this helped

5 0

3 years ago

A brown bear runs at a speed of 9.0\,\dfrac{\text m}{\text s}9.0 s m 9, point, 0, start fraction, start text, m, end text, div

KengaRu [80]

Explanation:

Below is an attachment containing the solution.

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3 years ago

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Can someone help me with this Physics question please?

strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

$N=N_0(\frac{1}{2})^{\frac{t}{H}$ where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

$N=48.0(\frac{1}{2})^{\frac{49.2}{12.3}$ and simplifying a bit:

$N=48.0(.5)^4$ and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

6 0

3 years ago

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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

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The center of the Milky Way galaxy lies in the direction of the _constellation, aboutlight-years away.

Elza [17]

Answer:

Sagittarius

Explanation:

The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation, about 26,000 light-years away.

5 0

4 years ago

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