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3 years ago

Abnormal protrusion of the eye out of the orbit is known as

Physics

1 answer:

kifflom [539]3 years ago

7 0

Answer:

Exophthalmos

Explanation:

Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.

It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.

It is commonly caused by trauma or swelling of eye surrounding tissues resulting from trauma.

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Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

What is characteristic of both sound waves and

REY [17]

"<span>(2) They transfer energy" is the best option from the list regarding the </span>characteristics of both sound waves and <span>electromagnetic waves, but these energies are different. </span>

5 0

4 years ago

Read 2 more answers

Jessie and Jaime complete a 5.0 km race. Each has a mass of 68 kg. Jessie runs the race at 15 km/h; Jaime walks it at 5 km/h. Ho

Leto [7]

The total metabolic energy used by each to complete the course is determined as 656.91 J.

<h3>Kinetic energy of Jessie and Jaime</h3>

The kinetic energy of Jessie and Jaime is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of Jaime
v is speed

15 km/h = 4.17 m/s

5 km/h = 1.39 m/s

K.E = ¹/₂(68)(4.17)² + ¹/₂(68)(1.39)²

K.E = 656.91 J

Thus, the total metabolic energy used by each to complete the course is determined as 656.91 J.

Learn more about kinetic energy here: brainly.com/question/25959744

5 0

2 years ago

A player throws a football 50.0 m at 61.0° north of west. what is the westward component of the displacement of the football?

-BARSIC- [3]

Answer: 24.24 m

Explanation:

A player throws football 50.0 m at 61° North of west. we will write this in terms of horizontal and vertical components.

Horizontal component: 50 cos 61° = 24.24 m which is westwards

Vertical component: 50 sin 61° = 43.73 m which towards North.

Refer to diagram below.

Thus, the westward component of displacement of the football is the horizontal component of the displacement = 24.24 m.

7 0

3 years ago

Which gas law states that the pressure of a gas decreases when volume is increased and the temperature is unchanged?

NeX [460]

Which gas law states that the pressure of a gas decreases when volume is increased and the temperature is unchanged?

A. The General Gas Law

B. The Law of Cubic Expansion

C. Charles' Law

D. Boyle's Law

4 0

3 years ago