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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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A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the sur
zubka84 [21]

Answer:

T = 5163.89 N

Explanation:

Newton's first law:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the car

W: Weight of the car : In vertical direction

FN : Normal force : perpendicular to the ramp

T :Tension force:  at angle of 31.0° above the surface of the ramp

Calculated of the Weight  of the car (W)

W = m*g   m: mass   g:acceleration due to gravity

W =   1130-kg* 9.8 m/s² = 11074 N

x-y weight components

Wx =  11074 N*sin 25.0° = 4680.07 N

Wy = 11074 N*cos 25.0° = 10036.45 N

x-y Tension components

Tx = T*cos 25.0°

Ty = T*sin 25.0°

Newton's first law:

∑Fx =0 Formula (1)

Tx-Wx = 0

T*cos 25.0° - 4680.07 = 0

T*cos 25.0° = 4680.07

T =  4680.07 / cos 25.0°

T = 5163.89 N

4 0
3 years ago
The variable ______________ describes how quickly something moves.
IrinaK [193]

it's up in Gogle trust me

8 0
3 years ago
A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
Harman [31]

Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

3 0
3 years ago
How does the latent heat of fusion of water help slow the decrease in air temperature,perhaps preventing temperatures from falli
love history [14]

Answer:

Water has the ability to release a large amount of energy during the freezing process. All of this energy is released into the air, resulting in a greater movement of air particles that will increase the air temperature. This heat will prevent air temperatures from drastically falling below 0ºC.

Explanation:

The latent heat of water occurs when the water is changing its physical state. In other words, when substances are changing their physical state, the amount of heat calculated for this change is called latent heat, and as we have already said, this heat is not related to the heat exchanges between two systems, but the change of state. physicist.

When water is changing from its liquid state to a solid state, we call latent heat latent heat from fusion (this is because fusion is the name we give to when liquid water is turning to ice). In the process of freezing the page, latent heat releases a large amount of energy into the air near the water. This energy agitates the air molecules, generating heat and preventing the air from reaching temperatures below 0ºC.

6 0
3 years ago
10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave
zhannawk [14.2K]

Answer:

To create a second harmonic the rope must vibrate at the frequency of 3 Hz

Explanation:

First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,

f₁ = v/2L

where,

v = speed of wave = 36 m/s

L = Length of rope = 12 m

f₁ = fundamental frequency

Therefore,

f₁ = (36 m/s)/2(12 m)

f₁ = 1.5 Hz

Now the frequency of nth harmonic is given in general, as:

fn = nf₁

where,

fn = frequency of nth harmonic

n = No. of Harmonic = 2

f₁ = fundamental frequency = 1.5 Hz

Therefore,

f₂ = (2)(1.5 Hz)

<u>f₂ = 3 Hz</u>

5 0
3 years ago
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