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MA_775_DIABLO [31]
4 years ago
13

How can humans increase the rate of endangered of extinct species

Physics
2 answers:
labwork [276]4 years ago
5 0

I would think A. Reduce Environmental Impacts.

If you build and populate more zoos, some of the endangered species might not be able to survive in captivity

If you introduce new predators they are going to kill all the endangered species and if you pass new laws about land use some people may kill the species

BARSIC [14]4 years ago
3 0

It's D. If you introduce new predators to an environment you will reduce a species population, and introducing a new predator can be done if humans release a predator that isn't usually in that environment. Plus, I took the test. :-)

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Which list of factors best supports the concept that they are "just right" for the planet earth to support life?
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?during the first few days of a fast, what energy source provides about 90% of the glucose needed to fuel the body?
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A 0.68 kg squirrel is resting on a branch 8 meters above the ground. What is the gravitational potential energy of a squirrel? A
ANEK [815]

Answer:

The gravitational potential energy of a squirrel is 53.312 J.

Explanation:

We have,

Mass of a squirrel is 0.68 kg

It is placed at a height of 8 m above the ground.

It is required to find the gravitational potential energy of a squirrel. It is possessed by an object due to its position. Its formula is given by :

E=mgh\\\\E=0.68\times 9.8\times 8\\\\E=53.312\ J

So, the gravitational potential energy of a squirrel is 53.312 J.

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a quickly moving house cat has 10 j of kinetic energy at speed v. at what speed will the cat have 20 j of kinetic energy
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3 years ago
A block of mass M on a horizontal surface is connected to the end of a massless spring of spring constant k. The block is pulled
slamgirl [31]

Answer:

Minimum coefficient of kinetic friction between the surface and the block is \mu_k=\frac{kx}{2Mg} .

Explanation:

Given:

Mass of the block = M

Spring constant = k

Distance pulled = x

According to the question:

<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed.  </em>

So,

From the FBD we can say that:

⇒ Normal force, N=Mg                                   <em>...equation(i)</em>

⇒ Elastic potential energy, PE = \frac{kx^2}{2}               <em>  ...equation (ii)</em>

⇒ Frictional force, f = \mu_kN                                <em> ...equation (iii)</em>

⇒ Plugging (i) in (iii).

⇒ f=\mu_kMg

Now,

⇒ As we know that the energy lost due to friction is equivalent to PE .

⇒ PE=fx                     <em>...considering PE as</em> mgh or f(x) .

   Arranging the equation.

⇒ \frac{kx^2}{2}=\mu_k Mg (x)

⇒ \frac{kx}{2}=\mu_k Mg                 <em>...eliminating x from both sides.</em>

⇒ \frac{kx}{2Mg}=\mu_k                    <em>...dividing both sides wit Mg.</em>

Minimum coefficient of kinetic friction between the surface and the block is \frac{kx}{2Mg}=\mu_k .

4 0
4 years ago
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