What happens to the density if the volume of an object decrease and the mass remains the same?

A typical adult human lung contains about 330 million tiny cavities called alveoli. Estimate the average diameter of a single al

Amanda [17]

Answer:

The average diameter of a single alveolus is 0.0222 cm.

Explanation:

Volume of the lung ,V= 1.9 L

$1 L = 1000 cm^3$

$1.9 L=1.9\times 1000 cm^3=1900 cm^3$

Number of alveoli in a human lung = $330\times 10^6$

Volume of single alveoli =v

$v\times 330\times 10^6=V$

$v=\frac{1900 cm^3}{330\times 10^6}$

$v=5.7575\times 10^{-6} cm^3$

The alveoli are spherical.

Radius of an alveolus = r

Volume of the sphere = $\frac{4}{3}\pi r^3$

$v=\frac{4}{3}\pi r^3$

$5.7575\times 10^{-6} cm^3=\frac{4}{3}\times 3.14\times r^3$

$r=0.0111 cm$

Diameter of the alveolus =d

d = 2r = 2 × 0.0111 cm = 0.0222 cm

The average diameter of a single alveolus is 0.0222 cm.

7 0

3 years ago

Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh

iragen [17]

For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

6 0

2 years ago

Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings

ozzi

Answer:

$0.3165\ \text{rad/s}$

Explanation:

m = Mass of person = 65 kg

d = Diameter of round table = 6.5 m

r = Radius = $\dfrac{d}{2}=3.25\ \text{m}$

v = Velocity of person running = 3.8 m/s

$I_t$ = Moment of inertia of turntable = $1850\ \text{kg m}^2$

Moment of inertia of the system is

$I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2$

As the angular momentum of the system is conserved we have

$L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}$

The angular velocity of the turntable is $0.3165\ \text{rad/s}$ .

3 0

3 years ago

One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resist

Svetach [21]

<h3>Answer:</h3>

(a) <u>i₁ = 0.03818 A = 38.18 mA</u>

(b) downward

(c) <u>i₂ = 0.01091 A = 10.91 mA</u>

(d) rightward

(e) <u>i₃ = 0.02727 A = 27.27 mA</u>

(f) leftward

(g) <u>Eₐ = 3.818 Volts</u>

<h3>Question:</h3>

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

<h3></h3><h3>Explanation:</h3>

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0

E₂ - i₃R₃ - i₁R₁ = 0

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

Eqn (2) implies:

12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

36 - 1200 i₃ = 6 - 100 i₃

1100 i₃ = 30

using this value in eqn of i₂:

i₂ = [6 - 100(0.02727)]/300

i₂ = (6 - 2.727)/300

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

<u></u>

(a)

(b)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>downward</u>

(c)

(d)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>rightward</u>

(e)

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>leftward</u>

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

<u>Eₐ = 3.818 Volts</u>

3 0

4 years ago

_________ is stored energy.

Ivan

Answer:

I think so is b kinetic o a potential

5 0

3 years ago