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Leya [2.2K]
1 year ago
12

A scuba diver fills her lungs to capacity (6.0 L) when 10.0 m below the surface of the water and begins to ascend to the surface

. Assume the density of the water in which she is swimming is 1000 kg/{eq}m^3 {/eq} and use g = 10 m/{eq}s^2 {/eq}
A. Assuming the temperature of the air in her lungs is constant, to what volume must her lungs expand when she reaches the surface of the water?
B. What effect would the warming of the air in her lungs have on the volume needed when she surfaces?
C. Assuming the temperature of the air in her lungs is constant, what effect does her ascent have on the vrms of the air molecules in her lungs?
Physics
1 answer:
Dimas [21]1 year ago
8 0

The lungs of a scuba diver should expand to a volume of 12 L, the volume of air in her lungs will grow with an increase in surface air temperature, and the rms speed of the air velocity in her lungs won't change.

It is stated that water has a density of 1000 kg/m3. The diver's 6L lung capacity allows her to descend 10m under the sea. We may suppose that the water's temperature is constant.

A. The pressure P₁ at the surface of water is 100000 Pa.

So, the pressure at the depth of 10m in water will be given by,

P₂ = 1000x10x10 + P₁

P₂ = 100000 + P₁

P₂ = 200000 Pa

Now, we can write,

P₁V₁ = P₂V₂

Putting values,

6 x 200000 = V₂ x 100000

V₂ = 12L.

Her lungs should thus enlarge to 12L.

B. Because of the continuous pressure at the surface, the air warming will further increase the amount of air in the lungs.

C. The Vrms are provided by,

Vrms = √(3RT/M)

The Vrms won't change if the temperature stays constant since the air's mass will remain constant.

To know more about Volume visit,

brainly.com/question/25736513

#SPJ4.

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Answer:

(a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

Explanation:

Given that,

Incidence angle = 26.7°

Refraction angle = 18.3°

(a). We need to calculate the refraction of liquid

Using Snell's law

n=\dfrac{\sin i}{\sin r}

Put the value into the formula

n=\dfrac{\sin 26.7}{\sin 18.3}

n=1.43

We need to critical angle for the liquid when surrounded by air

Using formula of critical angle

C=\sin^{-1}(\dfrac{1}{n})

Put the value into the formula

C=\sin^{-1}(\dfrac{1}{1.43})

C=44.37^{\circ}

(b). Given that,

Incidence angle = 37.5°

Speed of light in mineral v=2.17\times10^{8}\ m/s

We need to calculate the index of refraction

Using formula of index of refraction

n=\dfrac{c}{v}

Put the value into the formula

n=\dfrac{3\times10^{8}}{2.17\times10^{8}}

n=1.38

We need to calculate the angle of refraction

Using Snell's law

n=\dfrac{\sin i}{\sin r}

\sin r=\dfrac{\sin i}{n}

Put the value into the formula

\sin r=\dfrac{\sin 37.5}{1.38}

r=\sin^{-1}(\dfrac{\sin 37.5}{1.38})

r=26.17^{\circ}

Hence, (a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

3 0
3 years ago
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Niobium metal becomes a superconductor when cooled below 9K. Itssuperconductivity is destroyed when the surface magnetic fieldex
elena-14-01-66 [18.8K]

Answer:

the maximum current is 500 A

Explanation:

Given the data in the question;

the B field magnitude on the surface of the wire is;

B = μ₀i / 2πr

we are to determine the maximum current so we rearrange to find i

B2πr = μ₀i

i = B2πr / μ₀

given that;

diameter d = 2 mm = 0.002 m

radius = 0.002 / 2 = 0.001 m

B = 0.100 T

we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A

so we substitute

i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷

i = 500 A

Therefore, the maximum current is 500 A

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What is a Super Massive Black Hole?
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Answer:

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The tortoise and the hare are having a race over a course of length 0.52 km. The tortoise crawls along at a constant 0.075 m/s w
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Answer:the hare slept for 1 hr 55minutes.

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Speed=distance/time

Speed of tortoise is 0.075m/s=520m/t

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Answer:

0 (Zero)

Explanation:

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Work done on an object is the force applied on that object in the direction of Displacement of the object .

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So, W = 20×0 = 0.

Thus, Work done on the wall is zero.

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