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3 years ago

What are tiny sacs at the end of the bronchioles filled with air called?

Physics

1 answer:

WARRIOR [948]3 years ago

7 0

Answer:

Alveoli

The bronchioles end in tiny air sacs called alveoli, where oxygen is transferred from the inhaled air to the blood.

Hope this helps :)

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"How did your current and voltage measurements differ between the series and parallel circuits you created

irakobra [83]

Answer:

Series circuit:

The voltage that is measured across the circuit is different.

The current measured in a series circuit remains the same at all points in the circuit.

Parallel circuit:

The current measured across each resistor varies

The voltage measured across a parallel circuit will remain the same

Explanation:

Series and parallel circuits behave differently when it comes to the circulation of current and the interaction with a potential difference.

In a series circuit, the resistances are connected end to end. As a result, the voltage that is measured across the circuit is different once resistance is encountered. However, the current measured in a series circuit remains the same at all points in the circuit.

A parallel circuit behaves in an exactly opposite manner to the series circuit. In a parallel circuit, the resistances are connected side by side. As a result of this, the current measured across each resistor varies as there are circuit branches through which electric current can flow into. On the other hand, the voltage measured across a parallel circuit will remain the same

4 0

3 years ago

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

3 0

3 years ago

Which of these best explains the ability of small insects to walk on the surface of still water?

Furkat [3]

Well, the surface of still water has surface tension. If there isn't enough mass or weight to break the surface tension, the object will float.

7 0

3 years ago

Page 40-44 earth science regents<br> just post the picture of the pages please

Anna007 [38]

I don’t know what book you’re talking about so I can’t help but have a look online, you may be able to find it if you search up the book name and look around a few websites

8 0

3 years ago

You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when

jok3333 [9.3K]

U have to *modify it to increase its ground clearance*

4 0

3 years ago

Read 2 more answers