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Shkiper50 [21]
3 years ago
13

The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin

g the alveolus acts like a spherical bubble, what is the surface tension of the fluid and membrane around the outside of the alveolus? How does this compare to the surface tension of water?
Physics
1 answer:
alekssr [168]3 years ago
3 0

Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = 200\mu m = 200\times 10^{- 6}\ m

Gauge Pressure inside, P_{in} = 25\ mmHg

Blood Pressure outside, P_{o} = 10\ mmHg

Now,

Change in pressure, \Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

\Delta P = \frac{4\pi T}{R}

T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

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Vilka [71]

Answer:

(a)2.865 s

(b)2.865 s

Explanation:

We are given that

Acceleration,a=3.49 m/s^2

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Final speed,v=39 m/s

We know that

t=\frac{v-u}{a}

Using the formula

t=\frac{39-29}{3.49}=2.865 s

b.Initial speed,u=59 m/s

Final speed,v=69 m/s

Again using the formula

t=\frac{69-59}{3.49}=2.865 s

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2 years ago
Whats the difference between kinetic and potential energy?
damaskus [11]
Kinetic energy is energy in motion and potential energy is stored energy
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A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is att
ipn [44]

h =(3.7 - .58)m  = 3.12m

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√2 x 9.8 x 3.12

= 7.82m/s

I hope this helps!

7 0
2 years ago
An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

5 0
3 years ago
A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The
Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is W=2430\ pound

mass of roller coaster m=\frac{W}{g}=\frac{2430}{32.2}=75.45

Distance traveled by roller coaster d=396\ ft

drag force f_d=85\ pounds

velocity at top v=59 mph\approx 86.53\ ft/s

Suppose E is the initial energy

Conserving Energy at bottom and top

E=\frac{1}{2}mv^2+mgh+f_d\cdot d

E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396

E=2.9\times 10^5\ foot-pound

5 0
3 years ago
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