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Eva8 [605]
3 years ago
13

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 many significant figures are in 107 moles of sodium
Blizzard [7]

Answer:

3 significant figures

Explanation:

here  zero is counted as one of the significant figures since it lies next to an integer.

hence the number has 3 sfg

8 0
2 years ago
What body systems help these cells get the energy they need?
MakcuM [25]

The body systems that help cells get the energy they need are the digestive system and circulatory system. The first system is the digestive system because when you eat food, your body breaks it down using the stomach and other different parts of the digestive system.  The next one is the circulatory system because your blood carries the nutrients you acquired from food you digested to the cells throughout your body.

3 0
3 years ago
PLEASE HELP!! I’ll give brainliest pls
marin [14]

Answer:

A

Explanation:

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6 0
2 years ago
A 10 kg box is 1.3 m above the ground. How much potential energy does it have? (g on Earth of 9.8 m/s?
Volgvan
Potential energy = mgh
Potential energy = 10 x 9.8 x 1.3
Potential energy = 127.4 J
8 0
3 years ago
A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

7 0
3 years ago
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