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Reika [66]
3 years ago
6

What exhibits gill slits at some stage of its development?

Biology
1 answer:
IgorC [24]3 years ago
3 0
Animals in the Phylum Chordata. 
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A student does an experiment for a science fair to study whether temperature affects the timing of a cricket’s chirps. The stu
nexus9112 [7]

The frequency of chirps increases as the temperature increases. The correct option is B. <em>The chirps occur closer together as the temperature increases.  </em>

-------------------------------

<h2><u>Available data:</u></h2>

<h3>Data recorded by the student </h3>

<em><u>Day      Temperature   Average time between chirps (sec)</u></em>

1            21                             2. 5

2           22                            2. 6

3           23                            2. 2

4           24                            2. 3

5           25                            2. 0

6           26                             1. 8

7           27                              1. 9

8           28                             1. 8

9          29                             1. 4

10          30                             1. 2

11           31                              1. 5

<u>12          32                             1. 1</u>

In this experiment the student recorded the time in seconds <u>between </u><u>chirps</u>.

We can see that <em>as the </em><em>temperature increases</em><em>, in general, the</em><em> time</em><em> between</em><em> chirps decreases</em>.

  • At 21ºC the time between chirps is 2.5 seconds
  • At 26ºC the time between chirps is 1.8 seconds
  • At 32ºC the time between chirps is 1.1 seconds

According to this information, if the average time between chirps decreases with the temperature increase, we can assume that<em> the </em><em>frequency</em><em> of chirps </em><em>increases</em><em> as the temperature gets higher. </em>

<h3>Options,</h3>

<em>A) The higher the temperature, the fewer chirps there will be in 10 seconds</em>.

Wrong. The number of chirps in 10 seconds will depend, not only on the frequency but also on how long the chirps last. Since the student did not record the time of chirps, we can not make this conclusion.  

<em>B) The chirps occur closer together as the temperature increases</em>.

True. The chirps increase in frequency as the temperature increases, so they occur closer together.

<em>C) The chirps become farther apart as the temperature increases</em>.

Wrong. We can see how the time between chirps decreases as temperature increases, meaning that they are closer not farther.

D) There is no relationship between temperature and the time between chirps.

Wrong. If we make a graph we will see the tendency and the relationship between chirps and temperature.

Graphs usually explain the relationship between variables. In this example, the relationship would be inverse.

-------------------------------------

You can learn more about

at brainly.com/question/15210301

5 0
3 years ago
Mars had an orbital period of 1.88 years. In two or more complete sentences,explain how to calculate the average distance from m
Vanyuwa [196]
This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres

<span>Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
v=\frac{2\pi{r}}{T}
</span>=\frac{2\pi{r}}{1.88*365.25*86400} m/s, accounting for leap years
=3.371*10^{-8}\pi{r}   m/s

The centripetal force, Fc, generated is
Fc=\frac{mv^2}{r}   
        where m=mass of mars = 6.39*10^(24) kg
=\frac{mv^2}{r}
=\frac{6.39*10^{24}v^2}{r}
=7.26168*10^9\pi^2r

The gravitation pull from the sun, Fg, is given by
Fg=\frac{GMm}{r^2}
    where G=grav. const., =6.67408*10^(-11) m^3<span> kg^(</span>-1)<span> s^(</span>-2)
              M=mass of sun=1.989*10^(30) kg
=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}
=\frac{8.4826*10^44}{r^2}

Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}
Solving for the real root:
r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}
=\frac{1.1681263*10^{35}}{\pi^2}
=2.279*10^11 m


5 0
3 years ago
How would the increase of deforestation affect the amount of carbon in the atmosphere?
Flura [38]
First deforestation takes away trees which help absorb the carbon in the air and second if you burn the trees you release the carbon in the trees
6 0
3 years ago
As blood pressure increases and the hearts is working harder the most vulnerable arteries area? Located near the heart, because
KIM [24]

As blood pressure increases and the hearts is working harder the most vulnerable arteries area are those <u>Located near the heart, because they are absorbing the most shock.</u>

<u></u>

Explanation:

When the heart pumps blood, the highest pressure is felt close to the heart. This is why the aorta, the artery that carries blood away from the heart is made up of thick walls to withstand this pressure, otherwise, these vessels would rupture. Away from the heart, this pressure lessens, and the blood vessels are not so thick-walled.

Learn More:

For more on structure of arteries check out

brainly.com/question/2700868

brainly.com/question/1686670

#LearnWithBrainly

8 0
3 years ago
What is a scientific experiment?
astraxan [27]

Answer:

an experiment

Explanation: Any experiment you do to benefit the world in the name of science

3 0
3 years ago
Read 2 more answers
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