All categories

3 years ago

How do invasive species spread? identify three methods.Why do invasive species pose such a threat?

Physics

1 answer:

seropon [69]3 years ago

4 0

Answer:

See the explanation below.

Explanation:

A species is invasive can be spread, naturally either by the behavior of the species, or introduced accidentally or voluntarily by man.

The invasive species increases its density in the occupied area or colonizes new territories over time.

That an invasive species is harmful means that it produces significant changes in the composition, structure or processes of natural or semi-natural ecosystems, endangering native biological diversity.

Uncontrolled reproduction of this invasive species also increases the risk of control and extermination of the species, due to the acclimatization of the new species in the environment as well as possible exponential reproduction.

You might be interested in

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

The orbital speed is $v= 2.6*10^{3} m/s$

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago

Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Vedmedyk [2.9K]

Answer:

(a) V (A) = 0.7 m/s,

(b) V (A) = 0.7 m/s,

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a) law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V = 0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) = 0.7 m/s

b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) = 0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)

For elastic collision

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0

3 years ago

If a force of 10 n is applied to an object with a mass of 1kg the object will accelerate at

givi [52]

We Know, F = m*a
Here, F = 10 N
m = 1 Kg

Substitute their values in the equation,
10 = 1 * a
a = 10/1
a = 10

So, your final answer & the acceleration of the object would be 10 m/s²

Hope this helps!

3 0

3 years ago

The suns in the binary star system would be very far away, how does this

Vlad1618 [11]

It would impact the size by how far it is in km or light years

4 0

3 years ago

HELP PLS!!!!!!!! general science experiment : determining volume results

Vaselesa [24]

Answer:

this is the answer i gave and got a 95%

Explanation:

1. It depends what type of method your using. If it is Height * Width * Length then it will not work for irregular shaped objects because, it has extra pieces that wouldn't be there.

2. The second method would work for regular and irregular shapes because, you would have to find out the volume of the regular shape to get the volume for the irregular shape.

3. It depends on what your doing, if your doing a regular shape then use the first method, if it's an irregular shape then use the second method, if you do the math correctly both should give you an accurate answer for what you would like to achieve.

4. No, because the sugar would dissolve.

5. Not in this case the displacement tactic wasn't going to work becuase of weight

3 0

3 years ago