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raketka [301]
2 years ago
12

Lord howe island is a volcanic island in the tasman sea that is about 11 km long and 2. 8 km wide

Geography
2 answers:
mash [69]2 years ago
8 0

Lord howe island has two species of palm tree in Tasman sea which do not evolve in evolutionary process because Genetic Drift between the two species is not occurring. Thus the correct option is genetic drift of Howea forsteriana.

<h3>What is genetic drift?</h3>

Genetic drift is the random process which lead to large change in the population over a period of time. It is process which refers to the transfer of genes from population to the other due to migration or reproduction.

It is an evolutionary process which leads to change in the species which causes genetic variations. In Howea Forsteriana and Howea Belmoreana species interbreed do not take place because there is a lack of genes flow between them.

Thus genes flow is the process which is not taking place in the above scenario.

Learn more about volcanic island here:

brainly.com/question/1156480

#SPJ4

DedPeter [7]2 years ago
7 0

The evolutionary process that is not happening in the scenario mentioned is Hybridization.

<h3>What do you mean by Hybridization?</h3>

A method via which species can mate with every other and bring fertile offspring hence growing a brand new species is known as Hybridization.

This is not always going on with the two palm species on the island. This can be because of prezygotic boundaries that prevent those species from mating and generating a zygote (fertilized egg cell).

Therefore, In essence, those boundaries prevent any gene glide among the two species.

Learn more about Hybridization here:

brainly.com/question/14447584

#SPJ4

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Answer:

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Which continent would have the majority of the superimposed boundaries in the world?
zavuch27 [327]
Which continent would have the majority of the superimposed boundaries in the world?


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8 0
3 years ago
a rural kansas watershed that is ungauged has an area of 475 acres and a main channel length of 6870 feet with an average slope
dlinn [17]

Answer:

\mathbf{Q_p  =682 \ \  ft^3/s}

Explanation:

Given that:

Area = 475 acres

The length of the channel (L) = 6870 feet

The average water shield slope (S) = 100 feet/mile

Since; 1 mile = 5280 feet

Burst duration D = 15 min

∴

100 feet/mile = 100/5280

The average water shield slope (S) = 5/264

Using hydrograph method:

The time of concentration t_c = 0.0078L^{0.77} S^{-0.385}

where;

L = 6870

S = 5/264

t_c = 0.0078(6870)^{0.77} (\dfrac{5}{264})^{-0.385}

t_c =32.34 min

Since 60 min = 1 hour

32.34 min will be (32.34*1)/60

= 0.539 hour

Lag time T_l = 0.67\times t_c

T_l = 0.67\times 32.34

T_l = 21.6678\ min

The time to peak i.e

T_p = \dfrac{D}{2}+ T_L \\ \\  T_p = \dfrac{15}{2}+ 21.6678 \\ \\  T_p = 29.168 \ min

T_r = \dfrac{T_p}{5.5} \\ \\  T_r = \dfrac{29.1678}{5.5} \\ \\ T_r = 5.30 \ min

Since D = 15 min is not equal to T_r, then we hydrograph apart from T_r duration lag time.

Then;

T_p \ ' = T_p + \dfrac{D-t_r}{4} \\ \\ T_p \ ' = 29.168 + \dfrac{15-5.30}{4} \\ \\ T_p \ ' = 31.593

Now, we need to determine the peak discharge Q_p by using the formula:

Q_p  = \dfrac{484 \times A}{T_p \ '}

where

484 = peak factor

Recall that A = 475 acres, to miles, we have:

A = 0.7422 mile²

T_p \ ' = 31.593/60

∴

Q_p  = \dfrac{484 \times 0.7422}{\dfrac{31.593}{60}}

\mathbf{Q_p  =682 \ \  ft^3/s}

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irakobra [83]
I believe it is false 
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