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3 years ago

How is the surface area to volume ratio maximized in lungs?

1 answer:

8 0

There are branches on the lungs making it resemble a tree. These branches have sacs called alveoli that look like grape bunches. The combination of branching and alveoli maximize lung surface area.

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I need help with this answer

ololo11 [35]

The answer is Synthesis

7 0

3 years ago

A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20

alexandr402 [8]

Answer:

the amplitude of the subsequent oscillations is 0.11 m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider, $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of the subsequent oscillations is A = 0.11 m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and $v_{f}$ is the final speed.

to find $v_{f}$ , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

$(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}$

$v_{1}$ = 0, thus

(0.75+0.2) $v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11 m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π $\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π $\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s

7 0

4 years ago

Which bmi score listed below is considered over weight?

valentinak56 [21]

I don't see any answer choices, but if I remember correctly, the overweight BMI score is 20-22 and up. Probably 22 and up.

5 0

4 years ago

A horse walks along a straight path. it travels 3 in one second, 4 m in the next second, and 5 m in the third second. describe t

Temka [501]

Answer: The horse is moving with a uniform acceleration

Explanation:

According to the described situation, we are dealing with a constant acceleration (also called uniform acceleration), since the horse's velocity is changingn by a constant rate.

Let's prove it:

Firstly we are told the horse follows a straight path. In addition we are given its velocity:

3 m/s, then 4 m/s and 5 m/s

Since acceleration $a$ is defined as the change of velocity in time we can calculate it:

$a\frac{4 m/s-3 m/s}{1 s}=1m/s^{2}$ This is the horse's constant acceleration

3 0

3 years ago

Friction between two flat surfaces is distinguished as being either static friction or ________ friction

cricket20 [7]

Friction between two flat surfaces is distinguished as being kinetic friction.

3 0

3 years ago

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