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3 years ago

What is the relationship between the size of a predator population and the size of a prey population

1 answer:

3 0

<span>Let's put it this way. Say you have a killer-whale and a penguin. Killer-whales are major predators to penguins. Now, say the killer-whale population increases. The penguins would be eaten more by the killer-whales, then causing a population decrease for the penguins. If the population decreases, they're won't be enough penguins, and they most likely will become extinct, as well as causing a population decrease for the killer-whales as well. Whereas, vis versa, they're were a killer-whale population decrease. The penguins would be less hunted, therefore, creating a population increase for the penguins.</span>

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Which evidence taken from a crime scene would be the best evidence for a District Attorney to gain a conviction with DNA evidenc

mezya [45]

Answer:

Blood

Explanation:

All the mentioned option will yield the same DNA for the suspect but blood, besides the DNA also contain further information that can be matched to the victim like blood group, genotype, some antibodies present and even traces of substances and disease.

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3 years ago

If the refractive index of water is 1.33, then its critical angle is...

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2 years ago

A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

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3 years ago

What is the time required for an object starting from rest.

sashaice [31]

Time required for what?

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2 years ago

Never text and drive! Discuss with your parents the use of your cell phone when driving. Explain to them the one situation when

DanielleElmas [232]

One reason that it would be appropriate to talk on your cell phone, could be reporting an accident. For instance, a car crash, a sign in the road anything regarding safety of drivers.

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3 years ago