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Lostsunrise [7]
2 years ago
7

The term ���smog��� refers to the combination of:

Biology
1 answer:
Ivanshal [37]2 years ago
4 0
The answer is c. smoke and fog
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Which action best demonstrates the transformation of mechanical energy to heat energy? (HELP FAST)
vredina [299]

The action that BEST demonstrates the transformation of mechanical energy to heat energy is burning a candle (Option B).

Mechanical energy refers to the energy contained by an object due to its motion and/or position. Heat energy refers to the energy caused by the movement of atoms and/or molecules.

Energy transformation refers to the process by which the energy of an object changes from one type to another.

Mechanical energy can be converted into heat energy and vice-versa.

For example, thermal (heat) energy can be used to generate mechanical energy that produces the movement of a turbine generator, which subsequently converts mechanical energy into electrical energy.

Learn more in:

brainly.com/question/10555430?referrer=searchResults

8 0
1 year ago
I forgot the question
Bingel [31]

Answer: yes

Explanation: you forgot the question

7 0
3 years ago
Determine 3examples of surface water
vodomira [7]

Water...............

6 0
3 years ago
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Please help and be quick 5 ⭐️
amm1812

Answer:

Orbital

Explanation:

I believe, to the best of my knowledge, that the answer that you're looking for here is the Orbital. Hopefully this helps.

7 0
3 years ago
Read 2 more answers
In a population of cats, the phenotypic frequency of black cats is 91%, and the phenotypic frequency of white cats is 9%. Assumi
Talja [164]

Answer:

1. Allele frequency of b = 0.09 (or 9%)

2. Allele frequency of B = 0.91 (0.91%)

3. Genotype frequency of BB = 0.8281 (or 82.81%)

4. Genotype frequency of Bb = 0.1638 (or 16.38%)

Explanation:

Given that:

p = the frequency of the dominant allele (represented here by B)  = 0.91

q = the frequency of the recessive allele (represented here by b)  = 0.09

For a population in genetic equilibrium:

p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)

(p + q)^2 = 1

Therefore:

p^2 + 2pq + q^2 = 1

in which:  

p^2 = frequency of BB (homozygous dominant)

2pq = frequency of Bb (heterozygous)

q^2 = frequency of bb (homozygous recessive)

p^2 = 0.91^2 = 0.8281

2pq = 2(0.91)(0.9) = 0.1638

4 0
3 years ago
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