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4 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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Help. I need to find the degree using the sine law.

Marina CMI [18]

Use pythagorean theorem
$a ^{2} + b {}^{2} = c {}^{2}$
to find the opposite side, which is 7.3

so then you can just use inverse sinA=7.3/10 which equals 46.9 degrees

4 0

4 years ago

50200 J of heat are removed from

Dmitry_Shevchenko [17]

Correct Answer:

3.1375

Explanation:

Use equation $Q=mc$ Δ $T$ to find $m$

Plug in all variables $-50200=x\cdot 2000\cdot -8$

Answer: 3.1375

4 0

4 years ago

What is night vision pls explain in long answers use<br> can use google

bulgar [2K]

Answer:

Night vision is the ability to see in low-light conditions. Whether by biological or technological means, night vision is made possible by a combination of two approaches: sufficient spectral range, and sufficient intensity range.

8 0

3 years ago

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh

Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = $\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}$

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0

3 years ago

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

2 years ago