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seraphim [82]
3 years ago
12

How many lobes are in the lungs

Physics
1 answer:
wolverine [178]3 years ago
3 0

Answer:

There are 5 lobes in the lungs.

Explanation:

The right lung consists of three lobes: the RUL, the RML, and the RLL. While the left lung consists of two lobes: the LUL, and the LLL.

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Independent practice
coldgirl [10]

Explanation:

formula: <u>Mass</u>

Density x volume

2a) m=10kg v=0.3m³

10÷0.3=33.3 kg/m

2b) m = 160 kg V=0.1m³

160÷0.1=1600 kg/m

2c) m = 220 kg V = 0.02m³

220÷0.02=11000 kg/m

A wooden post has a volume of 0.025m³ and a mass of 20kg. Calculate its density in kg/m.

density = volume ÷ mass

20÷ 0.025=800 kg/m

Challenge: A rectangular concrete slab is 0.80m long, 0.60 m wide and 0.04m thick. Calculate its volume in m³.

Formula : Length x width x height = Volume

0.80 x 0.60 x 0.04 = 0.0192m³

B) The mass of the concrete slab is 180 kg. Calculate its density in kg/m.

density = volume ÷ mass

180 ÷ 0.0192 = 9375 kg/m

4 0
2 years ago
A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration
bazaltina [42]

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

3 0
3 years ago
Help with 3 and check the other ones please :(
Kobotan [32]

Answer:

number 3 The wind because the wind has strong pressure which causes to change peoples directions in where they are going

Explanation:

hope this helped by the way i think you answered b and 4 i don't know if this is a test and i sent this at the wrong time :(

4 0
3 years ago
Read 2 more answers
What do radio waves and microwaves have in Common?
DerKrebs [107]

What do radio waves and microwaves have in Common?

<u>A) Both are at the ideological the spectrum that has the lowest frequency. </u>

B) Both are at the spectrum that has the shortest wavelengths.

C) Both have higher frequencies than visible light.

3 0
3 years ago
A 1250-kg compact car is moving with velocity v1 =36.2i^+12.7j^m/s. It skids on a frictionless icy patch and collides with a 448
MA_775_DIABLO [31]

Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:

(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>

where <em>v</em> is the velocity of the system. Solve for <em>v</em> :

<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)

<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s

7 0
2 years ago
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