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EleoNora [17]
3 years ago
5

Population growth is limited by density-dependent factor such as____

Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0

Answer:

predation, water supply, parasitism, migration, diseases, waste accumulation, food availability, and competition.

Explanation:

A population can be defined as the total number of living organisms living together in a particular place and sharing certain characteristics in common.

Population regulation can be defined as a biological process that balances limiting factors affecting the growth of a population based on density.

Basically, the factors that regulate the growth of a population are divided into two (2) main categories and these includes;

I. Density-independent factors.

II. Density-dependent factors.

Density-dependent are regulating factors such as predation, diseases, and competition that affect the size of the population of living organisms through decreasing or increasing mortality and birth rate.

This ultimately implies that, population growth is limited by density-dependent factors such as predation, parasitism, migration, diseases, waste accumulation, food availability, water supply, and competition within the population.

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P= mv
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7 0
3 years ago
F. If the shuttle's period is synchronized with that of Earth's rotation, what is the height of the shuttle? (1 day = 8.64x104s,
oee [108]

Answer:

1.324 × 10⁷ m

Explanation:

The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.

Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.

We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have

Rω² = GME/R²

R(2π/T)² = GME/R²

R³ = GME(T/2π)²

R = ∛(GME)(T/2π)²

RE + h = ∛(GMET²/4π²)

h = ∛(GMET²/4π²) - RE

substituting the values of the variables, we have

h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m

h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m

h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m

h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m

h = 19.64 × 10⁶ m - 6.4 × 10⁶ m

h = 13.24 × 10⁶ m

h = 1.324 × 10⁷ m

3 0
3 years ago
How do I use a metric conversions chart
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4 0
4 years ago
If hydrogen and helium are the most abundant elements in the unoverse , then why are living organisms composed primarily of carb
Mumz [18]

Answer:

Most of materia isnt life.

Explanation:

The living organisms (life) aren't the most abundant thing in universe.

Hydrogen and helium are present in everywhere, but life isn't.

There is no reason to think because we have a lot of a thing, the life must be made for this thing.

The organic life just can exists because some mysterious properties about carbon, that is the basic foundation of life, carbon is a special element, why? We don't know, actually, it's a huge problem for science discover why the carbon can makes life be possible and other elements can't. But we know is this element that makes life possible.

So, note there isn't relation about the quantity of a material in Universe and the life constituition. In addition, look around, organic materials are very rare in Universe, Earth is one in lots of places and in most of this places there isn't sign of life.

Even in Earth the life looks abundant, in Universe it isn't, the same way in Universe the Hydrogen and Helium are abudant, in Earth isn't soo.

3 0
3 years ago
A person standing 1.20 m from a portable speaker hears its sound at an intensity of 5.50 ✕ 10−3 W/m2. HINT (a) Find the correspo
iris [78.8K]

Answer:

PART A)

L = 97.4 dB

PART B)

I = 6.11 \times 10^{-6} W/m^2

PART C)

L = 67.9 dB

Explanation:

PART A)

level of sound is given as

L = 10 Log\frac{I}{I_o}

now we have

L = 10 Log\frac{5.50\times 10^{-3}}{10^{-12}}

L = 97.4 dB

PART B)

Since source is a spherical source

so here the intensity of sound is inversely depends on the square of the distance from the source

\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}

\frac{I_2}{5.50 \times 10^-3} = \frac{1.20^2}{36^2}

I_2 = 6.11 \times 10^{-6} W/m^2

PART C)

level of sound is given as

L = 10 Log\frac{I}{I_o}

now we have

L = 10 Log\frac{6.11\times 10^{-6}}{10^{-12}}

L = 67.9 dB

3 0
3 years ago
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