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Sergio039 [100]
3 years ago
12

Describe the circumstances of a solar eclipse.

Biology
1 answer:
dangina [55]3 years ago
3 0

A solar eclipse happens when the Moon passes in front of the Sun and casts a shadow across Earth.

For a total eclipse to take place, the sun, moon and Earth must be in a direct line. The second type of solar eclipse is a partial solar eclipse. This happens when the sun, moon and Earth are not exactly lined up. The sun appears to have a dark shadow on only a small part of its surface.

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The science that applies the mechanical principles of physics and engineering to the investigation of human movement and the str
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Biomechanics

Is the study of human movement
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Which of the following is required for photosynthesis to take place? a. sugar b. oxygen c. atp d. water please select the best a
levacccp [35]

The answer will be :

D. water.

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2 years ago
Cow's milk should never be fed to infants because it is too high in protein and too low in sodium. select one:
TEA [102]
Answer is B: false

it's too high in protein and too low in iron
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3 years ago
Two states of matter present in the water cycle are described below.
mars1129 [50]

State 1 is liquid cause liquids dont have a fixed shape and are free to take the shape of the container that it is in. State 2 is a solid because solids have a fixed shape and arent able to take the shape of the container

7 0
3 years ago
Albinism is an autosomal recessive trait in humans. Assume that there are 100 albinos (aa) in a population of 1 million. How man
andreyandreev [35.5K]

Answer:

980,100

Explanation:

let the recessive condition (aa) be represented as q² which is the genotypic frequency = 100/1,000,000 = 0.0001

allelic frequency of a = q = √q² = √0.0001 = 0.01. thus q = 0.01

From the formula p + q = 1 where p is the allelic frequency of A.

since q = 0.01

p = 1 - 0.01 = 0.99. The allelic frequency of A (p) = 0.99

p² = 0.99² = 0.9801. Genotypic frequency of AA= 0.9801

= 0.9801 x 1,000,000 = 980,100 individuals with AA (homozygous normal)

For 2pq genotypic frequency for hetrozygous (Aa).

Using the formula p² + 2pq + q² = 1.  Since p² = 0.9801 and q² = 0.0001

   2pq = 1 - (p² + q²)

= 1 - (0.9801 + 0.0001)

= 1 - (0.9802)

= 0.0198 = 0.0198 x 1,000,000 = 19,800 individuals with Aa

7 0
3 years ago
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