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DanielleElmas [232]
3 years ago
10

Why is there so much variation in human skin coloration? a. This occurred because humans underwent natural selection, which resu

lted in traits that conferred advantages to humans becoming fixed in populations. b. This occurred because alleles that are harmful in one environmental context may be beneficial in another. c. This occurred because humans are found in areas that range from high-UV-light environments to low-UV-light environments. d. This occurred because skin coloration results in different advantages depending on the levels of sun in the region.e. All of these.
Physics
1 answer:
Arlecino [84]3 years ago
5 0

Answer:

<u>e. All of these.</u>

Explanation:

  • The skin in humans ranges from darkest to the lightest color, as a result of genetic makeup, exposure to the sun rays, skin pigmentation may be either due to the evolutionary process of natural selection, and it may be due to the biochemical effects of the UV rays.
  • As the pigment in the skin of humans is affected by the content of melanin in the body that causes the determination of skins cells of the darker colored humans, and the light skin is determined by the bluish-white tissues under the dermis and the hemoglobin.
  • The emergence of skin pigments dates back to 1.2 billion years ago. When the harsh climatic conditions drove the early humans into arid and open landscapes. In general, the people living near to equator have darkly pigmented than those living in poles are lightly pigmented.
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An outdoor Wi-Fi unit for a picnic area has a 110 mW output and a range of about 38 m. What output power (in mW) would reduce it
valkas [14]

Answer:

The output power is 24.68 mW.

Explanation:

Given that,

Power = 110 mW

Range = 38 m

Reduced range = 18 m

We need to calculate the power

Using formula of intensity

I=\dfrac{P}{A}

I=\dfrac{P}{\pi r^2}

As intensity is constant

P\propto r^2

So, \dfrac{P_{1}}{P_{2}}=\dfrac{r_{1}^2}{r_{2}^{2}}

P_{2}=\dfrac{r_{2}^2}{r_{1}^{2}}\timesP_{1}

Put the value into the formula

P_{2}=\dfrac{18^2}{38^2}\times110\times10^{-3}

P_{2}=24.68\times10^{-3}\ W

P_{2}=24.68\ mW

Hence, The output power is 24.68 mW.

8 0
4 years ago
A capacitance C and an inductance L are operated at the same angular frequency.
PIT_PIT [208]

Answer:

(B) 7279.70 rad/sec (C) X_L=37.1265\ ohm\ and \ X_C=37.1265\ ohm

Explanation:

We have given L=5.10mH and C=3.70\mu F

(B) The angular frequency is denoted as \omega whose value is given by \omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.10\times 10^{-3}\times 3.70\times 10^{-6}}}=7279.70\ rad/sec

(C) The inductive reactance X_L=\omega L=7279.70\times 5.10\times 10^{-3}=37.1265\ ohm

Capacitive reactance  X_C=\frac{1}{\omega C}=\frac{1}{7279.70\times 3.70\times 10^{-6}}=37.1265\ ohm

6 0
3 years ago
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
Ludmilka [50]

Answer:

Force applied to stop the car = 1,250 N

Explanation:

Given:

Mass of car (M) = 1,000 kg

Initial velocity (U) = 20 m/s

Final velocity (V) = 0 m/s

Distance (S) = 160 m

Find:

Force applied to stop the car.

Computation:

v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N

Force applied to stop the car = 1,250 N

7 0
3 years ago
From a penalty kick, the ball rebounds off the goalkeeper back to the player who took the kick. That player then kicks the ball
Alexxx [7]

The correct restart is for there to be another kick off taken in this type of scenario.

<h3>What is Kick-off in Soccer?</h3>

This is a method of restarting play in which the ball is put on the center circle and passed by a player.

Kick offs occur at the beginning of any halves or when a goal is scored during the match which is why it's the most appropriate choice.

Read more about Soccer here brainly.com/question/12597997

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2 years ago
Which situations might cause two observers (A and B) to measure different frequencies for the same vibrating object? Select the
Alex787 [66]

We want to explain why two different observes may measure different frequencies for the same vibrating object.

We will see that the two correct options are:

  • <em>Observer A is stationary and Observer B is moving.</em>
  • <em>Observer A and Observer B are moving at different speeds relative to each other.</em>

<em />

Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

Now there appears a really cool effect, called the Doppler Effect. It says that the apparent change of frequency is <u>due to the motion of the observer or the source of the frequency (or both).</u>

For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

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If you want to learn more, you can read:

brainly.com/question/17107808

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