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3 years ago

Countercurrent exchange is evident in the flow of _____.

Physics

1 answer:

Anna35 [415]3 years ago

4 0

Answer:

In the water flow mechanism in fishes when they take in oxygen using the gills

Explanation:

Fish gills are organ that facilites respiration under water,its Mechanism is described below,first water is pulled in the mouth of the fish and pumped over the gills,while the blood flow in opposite direction known as counter current exchange helps in pushing low oxgen water away and retaining high oxgen in the fish blood.

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Of the three sciences, physics, chemistry, and biology, the most complex is

Agata [3.3K]

Answer:

I would probably say C to be completely honest

Explanation:

If you agree make sure to give me a like

6 0

2 years ago

Does the sun make Gamma rays?<br>what gives Infrared light?

Gnesinka [82]

Sun fives off both of them

4 0

3 years ago

Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0

Inessa05 [86]

Let $a_1$ be the average acceleration over the first 2.46 seconds, and $a_2$ the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let $v$ be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

$a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}$

and we're also told that

$\dfrac{a_1}{a_2}=1.66$

(or possibly the other way around; I'll consider that case later). We can solve for $a_1$ in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

$1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

Now we can solve for $v$ . We find that

$v=20.8\,\dfrac{\mathrm m}{\mathrm s}$

In the case that the ratio of accelerations is actually

$\dfrac{a_2}{a_1}=1.66$

we would instead have

$\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)$

in which case we would get a velocity of

$v=24.4\,\dfrac{\mathrm m}{\mathrm s}$

6 0

3 years ago

Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman

Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2$