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Alex777 [14]
3 years ago
14

Green sea turtles eat sea grass that grows on the sea floor. Seagrass, a foundation species, provides an important breeding grou

nd for fish and other marine organisms. A conservation program for the green sea turtles plans to introduce more turtles into an ecosystem. What could be the immediate effect of this on the ecosystem?
A) More species of fish would migrate to the region.
B) The new turtles would drive out the existing population.
C) The region would witness a spurt in the growth of sea grass.
Eliminate
D) There would be a decrease in the population of marine organisms.
Physics
2 answers:
d1i1m1o1n [39]3 years ago
3 0
The answer is D. <span> There would be a decrease in the population of marine organisms. </span>
Agata [3.3K]3 years ago
3 0

The answer is D.  There would be a decrease in the population of marine organisms.


Populations are effected not only because of predation but also due to habitat loss caused by the introduction of a new species into the ecosystem. If more sea turtles are added into the region they would deplete the sea grass, causing a decrease in the population of marine organisms due to the loss of breeding grounds and food source.

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Answer:

1050 kg

Explanation:

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KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.

We can now substitute the values we know into this equation.

KE = 472 500 J and v = 30 m/s:

472 500 = 1/2 × m × 30²

Next, we can rearrange the equation to make m the subject and solve for m:

m = 472 500 ÷ (1/2 × 30²)

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2 years ago
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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
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Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

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Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

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The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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