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3 years ago

Please help me do this problem

Physics

1 answer:

Digiron [165]3 years ago

4 0

Answer:

i) acceleration from B to D is 0, because the velocity is constant (stays the same)

ii) whatever units of distathat might be, we can calculate the number:

for 4 time-steps (2 to 6) the velocity is 6 per time step, that makes 24 distance units in these 4 time steps. it's the same the area underneath the graph.

there is also the vertical line from 0 to 2. we can calculate that distance like the area of a triangle with 2*6 / 2 = 6

the total distance from 0 to D is therefore 30

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The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in

mihalych1998 [28]

Answer:

a) Ra = 0.517 Ω

Rb = 0.032 Ω

Rc = 0.129 Ω

b) Ia = 5.8A

Ib = 93.75A

Ic = 23.2 A

Explanation:

a) The resistance is equal to:

Resistance for case a:

$R_{a} =\frac{pL_{a} }{A_{a} } =\frac{p*4*L_{0} }{2L_{0}*L_{0} } =\frac{2p}{L_{0} }$

Where

p = 1.5x10⁻²Ωm

L0 = 5.8 cm = 0.058 m

$R_{a} =\frac{2*1.5x10^{-2} }{0.058} =0.517ohm$

Resistance for case b:

$R_{b} =\frac{pL_{b} }{A_{b} } =\frac{pL_{0}}{2L_{0}4L_{0} } =\frac{p}{8L_{0}} =\frac{1.5x10^{-2} }{8*0.058} =0.032ohm$

Resistance for case c:

$R_{c} =\frac{pL_{c}}{A_{c} } =\frac{p2L_{0}}{L_{0}4L_{0}} =\frac{p}{2L_{0}} =\frac{1.5x10^{-2} }{2*0.058} =0.129ohm$

b) The current is equal to:

Current for case a:

$I_{a} =\frac{V}{R_{a} } =\frac{3}{0.517} =5.8A$

Current for case b:

$I_{b} =\frac{V}{R_{b} } =\frac{3}{0.032} =93.75A$

Current for case c:

$I_{c} =\frac{V}{R_{c} } =\frac{3}{0.129} =23.2A$

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3 years ago

Assortative mating changes ______ frequencies but does not change ______ frequencies.

andrew-mc [135]

Allele frequencies are unaffected by assortative mating, but genotype frequencies .

<h3>Assortative mating: </h3>

Individuals with similar phenotypes and genotypes mate with others more frequently than is anticipated under a random mating pattern in assortative mating, which is a mating pattern and a type of sexual selection.

<h3>Frequencies of genotypes:</h3>

A population's genotype frequency is calculated by dividing the number of people having a particular genotype by the overall population size. The genotype frequency in population genetics is the frequency or ratio (i.e., 0 f 1) among genotypes inside a population.

<h3>The frequency for alleles in biology:</h3>

The term "allele frequency" describes the prevalence of an allele in a population. It is calculated by calculating the number of times the allele occurs in the population and dividing by the sum of all the gene copies.

To know more about Assortative mating visit:

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