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alexira [117]
3 years ago
11

How much work is done when a force of 15.0 newtons is used to lift a box 3.0 meters into the air?

Biology
2 answers:
enyata [817]3 years ago
5 0
<h3><u>Answer;</u></h3>

45 joules

<h3><u>Explanation;</u></h3>

Work done by an object is given by the product of force and distance.

Work done is measured in Joules.

<em>Work done = Force × distance;</em>

<em>Force = 15 Newtons, and Distance = 3.0 meters </em>

Therefore;

<em>Work done = 15 × 3</em>

                   <u>= 45 Joules </u>

Annette [7]3 years ago
4 0
W= f x d

W= 15 x 3

W= 45J
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In a species of bird, the allele for the striped trait is recessive to the allele for the speckled trait. Suppose that a coastal
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The answer is "0.42"

Explanation:

Please find the complete question in the attached file.

Coastal community striped allele intensity = q = 0.62

Landlocked community, strip intensity allele=q'=0.40

Its eliminated allele frequency for movement = q"=0.42 \ in \ indoorscommunity And now in the sturdy vineyard optimum communities, genotype frequency of stripped characteristic (heterozygous device) in coasts=q_2=0.39

Heritable (striped) allele frecency:

\to q=\sqrt{q_2}

      =\sqrt{0.39} \\\\ =0.624 \\\\ =0.62

(round up to the closest cent) is therefore the only frequency of coastal people.

The genome of pulling function in host population= q'_2=0.16 Heterozygous feature regularly.

Therefore the inland community rate of recessed (striped), allele rate:

\to q'= \sqrt{q'_{2}}= \sqrt{0.16}=0.40

Following migration;

Its percentage of coastal migrants:

m=10\% \\\\

   = \frac{10}{100} \\\\= 0.1

Coastal population Non-immigrant percentage of coastal residents:

=1- m\\\\=1-0.1\\\\ =0.9

Stripping coastal community of allele rate after immigration:

(q")=\text{(stripping of migrant only freq)}\times (m)+\text{(stripping allele baud rate for non-immigrants)} \times ( 1-m)

       = [q \times m]+[[q' \times(1-m)]\\\\=(0.62 \times 0.1)+(0.40 \times 0.9)\\\\=0.062+ 0.36\\\\=0.422\\\\=0.42

8 0
3 years ago
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