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svp [43]
2 years ago
7

According to the evolutionary perspective of mate selection, why do men and women exhibit different preferences in mate selectio

n? a. each gender has differing hormone compositions. b. each gender has differing societal expectations and restrictions. c. each gender has differing attraction cues and priorities. d. each gender has differing innate social capabilities. please select the best answer from the choices provided a b c d
Physics
1 answer:
Alla [95]2 years ago
6 0

C. Each gender has differing attraction cues and priorities.

<h3>Why do men and women exhibit different preferences in mate selection?</h3>

Men and women exhibit different preferences in mate selection because both have different attraction cues and priorities for their partner. These cues and priorities leads to difference in preference of life partner.

So we can conclude that Each gender has differing attraction cues and priorities.

Learn more about mate here: brainly.com/question/25261401

#SPJ1

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Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

F_{n} = 30.11N

Direction of the net force (β)

\beta =tan^{-1} (\frac{29.13}{7.62} )

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

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The normal force acts to counter the gravitational force, that is the upward direction.
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7 0
3 years ago
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Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

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(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

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t = \frac{-4\times 28.5}{log\frac{1}{2}}

t = 387.69 yrs

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