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3 years ago

How are limiting factors related to carrying capacity

1 answer:

8 0

Limiting factors are resources or other factors in the environment that can lower the population growth rate. ... The carrying capacity (K) is the maximum population size that can be supported in a particular area without destroying the habitat. Limiting factors determine the carrying capacity of a population.

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Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum

marta [7]

Answer:

$v_{top} = 61.96 \frac{mi}{h}$

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

$v_{top} = 27.7 \frac{m}{s}$

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

$1609.34 \ m = 1 \ mi$

Now, we can divide by 1609.34 meters on both sides:

$\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}$

The left sides equals 1, so

$1 = \frac{1 \ mi}{ 1609.34 \ m}$

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

$v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}$

$v_{top} = 0.0172120 \frac{mi}{s}$

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

$1 \ h = 3600 \ s$

as the seconds are dividing in the velocity, we know divide by 1 hour.

$\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}$

$1 = \frac{3600 \ s}{ 1 \ h}$

and know we just multiply our top speed by this conversion factor

$v_{top} = 0.0172120 \frac{mi}{s} \frac{3600 \ s}{ 1 \ h}$

$v_{top} = 61.96 \frac{mi}{h}$

8 0

3 years ago

n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

2 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago

An airplane pilot flies due west at a speed of 216 km/hr with respect to the air. After flying for a half an hour, the pilot fin

Yuki888 [10]

Answer:

speed wind Vw = 54.04 km / h θ = 87.9º

Explanation:

We have a speed vector composition exercise

In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south

Let's add the vectors in each coordinate axis

X axis (East-West)

-Xvion - Xw = -119

Xw = -Xavion + 119

Xw = 119 -108

Xwi = 1 km

Calculate the speed for time of t = 0.5 h

Vwx = Xw / t

Vwx= 1 /0.5

Vwx = - 2 km / h

Y Axis (North-South)

Y plane - Yi = -27

Y plane = 0

Yw = 27 km

Vwy = 27 /0.5

Vwy = 54 km / h

Let's use the Pythagorean theorem and trigonometry to compose the answer

Vw = √ (Vwx² + Vwy²)

Vw = R 2² + 54²

Vw = 54.04 km / h

tan θ = Vwy / Vwx

tan θ = 54/2 = 27

θ = Tan⁻¹ 1 27

θ = 87.9º

The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º west from the south

5 0

3 years ago

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rjkz [21]

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5 0

3 years ago

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