All categories

2 years ago

What is the FUNCTION of cilliated

Physics

1 answer:

Dahasolnce [82]2 years ago

5 0

Explanation:

In the trachea's inner layer, you have small, hair-like structures called cilia. Cilia move in rhythm to push mucus out of your trachea so that you either expel or swallow it

You might be interested in

Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

Eva8 [605]

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

$vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2$

Replacing:

$Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9$ $Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9$ $Vf1=-1-2.6=-3.6\text{ m/s}$

Answer: 3.6 m/s west

6 0

1 year ago

Rudy plans to conduct an experiment using three rosebushes of the same variety and size. His hypothesis is that the plant

Gala2k [10]

Answer: B) He has too many independent variables.

Explanation: Nothing in the experiment is a constant or being measured.

5 0

3 years ago

How do scientists believe the universe formed?

Arte-miy333 [17]

Answer:

The Big Bang was the moment that the universe began, that as a tiny dense and the exploded!!!!!!! BOMBBBBBBBBBBBBBBB

Explanation:

3 0

3 years ago

The study of origins and life of the planet earth is called

Luden [163]

Abiogenesis is the name of life on Earth.

8 0

4 years ago

(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current

erica [24]

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

$P=VI$

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

$P=(20 V)(18 A)=360 W$

(c) 360 J

The energy dissipated by the internal resistance is given by

$E=Pt$

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

$E=(360 W)(1 s)=360 J$

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

$P=I^2 R$

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$

4 0

3 years ago