All categories

3 years ago

What happens if actions taken during the fragmentation stage are only partially successful?

Physics

1 answer:

gogolik [260]3 years ago

3 0

Answer:

The community may regress to the denial stage or stagnate in the fragmentation stage.

Explanation:

An habitat is a place where organisms live. It provides the basic essentials of living for the organisms. Examples are; terrestrial, aquatic etc. Fragmentation of an habitat can occur due to some natural or artificial processes.

Habitat fragmentation is the process in which a large area of habitat is transformed into a number of smaller areas. These smaller patches are segregated from each other.

This may cause the extinction of some species due to edge effect.

If actions taken during the fragmentation stage are partially successful, the community in the habitat may regress to the denial stage or stagnate in the fragmentation stage.

You might be interested in

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

3 years ago

A 5.0 kg chunk of putty moving at 10m/s collides and sticks to a 7.0 kg bowling ball that is initially at rest.What is the total

Flura [38]

Answer:

Total momentum = 50kgm/s

Explanation:

Given the following data;

Mass, M1 = 5kg

Mass, M2 = 7kg

Velocity, V1 = 10m/s

Velocity, V2 = 0m/s (since it's at rest).

To find the total momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = Mass * Velocity$

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

Total momentum = M1V1 + M2V2

Substituting into the equation, we have;

Total momentum = 5*10 + 7*0

Total momentum = 50 + 0

Total momentum = 50 kgm/s

Therefore, the total momentum of the bowling ball and the putty after they collide is 50 kgm/s.

8 0

3 years ago

An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?

anyanavicka [17]

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

8 0

3 years ago

Read 2 more answers

Humpback whales are known to produce a collection of elaborate and repeating sounds with frequencies starting at 20 Hz. The soun

jasenka [17]

Answer:

70 m.

Explanation:

Given,

Frequency, f = 20 HZ

speed of sound, v = 1400 m/s

wavelength of the waves = ?

we know,

v = f λ

$\lambda= \dfrac{v}{f}$

$\lambda= \dfrac{1400}{20}$

$\lambda=70\ m$

Hence, the wavelength of the wave is equal to 70 m.

8 0

3 years ago

Definition of complex machine

ASHA 777 [7]

A complex machine is a machine made up of two or more simple machines that make your work easier to do. There are six simple machines from which all complex machines are made. They include: The lever. The inclined plane

4 0

3 years ago

Read 2 more answers