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2 years ago

Who discovered the advantages of AC over DC( alternating current and direct current)?

Biology

1 answer:

stira [4]2 years ago

8 0

Answer:

At the close of the nineteenth century, Thomas Edison was working on perfecting his system of direct current (DC) for electric power distribution. Meanwhile, the inventor and physicist Nikola Tesla was discovering the benefits of his competing alternating current (AC) system.

Explanation:

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Deep-ocean waters flow in an overturning circulation called _______

Marrrta [24]

The answer is currents

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3 years ago

What does the line of blue triangles mean on a weather map?

Fantom [35]

The blue triangles mean a cold front

3 0

2 years ago

Please help ASAP, thank you

Nesterboy [21]

Answer:

Explanation:

7 0

4 years ago

Read 2 more answers

A flower with reduced or inconspicuous sepals and petals and a large, feathery stigma most likely is pollinated by_________ . Co

rosijanka [135]

Answer:

Wind, Insects/animals

Explanation:

Pollination is the process by which pollen grains move from the anther(male part) to the stigma( female part).

Pollination by wind is usually made possible with flowers who have feathery stigma. The feathery stigma is easily swayed by the wind which makes it possible for it to catch pollen grains in the air.

Pollination by insects/animals is made

Possible by a form of attraction which is usually in the form of conspicuous petals and a good odor. These features attracts the insects/animals to the flowers.

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4 years ago

Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A 2 A 2) is 0.0004

Simora [160]

Hardy Weinberg Equilibrium

Explanation:

Hardy Weinberg Equilibrium formula for allelic frequency is: $p+q=1$

Hardy Weinberg Equilibrium formula for genotypic frequency is: $p^2+q^2+2pq=1$

Here, $p$ =frequency of dominant allele

$q$ =frequency of recessive allele

$p^2$ =frequency of dominant genotype

$q^2$ =frequency of recessive genotype

$2pq$ =frequency of heterozygous genotype

frequency of affected individuals=0.0004(given)

here,affected individuals means the one's having Cystic Fibrosis and since it is recessive disease so we can write frequency of recessive genotype as: $q^2$ =0.0004

If $q^2$ =0.0004 $q$ will be 0.02;frequency of recessive allele will be 0.02

Using the formula $p+q=1$

$p+0.02=1\\p=1-0.02=0.98$

hence,frequency of dominant allele will be 0.98

Frequency of dominant genotype will be $p^2=(0.98)^2=0.9604$
Frequency of heterozygous genotype will be $2pq=2*0.98*0.02=0.0392$

7 0

4 years ago