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
The average velocity of a car that travels 450 km north in 9.0 h is 5.0 x 10 km/h
Answer:
63
Explanation:
You first have to add all the numbers together.
22+72+79+72+70 = 315
You divide the total by the amount of numbers (5)
315/5 = 63
The mean is 63
Answer:

Explanation:
We know that the frequency of the nth harmonic is given by
, where
is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

So the tension is:

Which for our values is:

Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀