PART B: Which of the following quotes best supports the answer to Part A? A “Everyone, like myself, was probably on the verge of fetching a rope, or asking where we could find a ladder, but then we looked around at each other and it was decided.” ( Paragraph 2) B “I remember that we were still full of games and laughter when we called down to him. He heard us shouting while we were playing” ( Paragraph 3) C “At first afraid to disobey the voice from the man in the well, we turned around and actually began to walk toward the nearest house, which was Arthur’s. But along the way we slowed down” ( Paragraph 5) D “As we were rising quietly and looking at the dark shadows of the trees we had to move through to reach our homes, he said, ‘Why didn’t you tell anyone?’ He coughed. ‘Didn’t you want to tell anyone?’” ( Paragraph 56)

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

Answer:

Time rate = 0.010026 kg/m³.s

Explanation:

We are given the following;

Total lung capacity; V = 6000mL = 6 x 10^(-3) m³

diameter of the trachea is 18mm = 0.018m

Air density; ρ = 1.225kg/m³

Velocity; v = 20cm/s = 0.20m/s

From the question, lung volume is decreasing at a rate of 100mL/s.

Thus, dv /dt = -100mL = -0.0001 m³/s

Area of trachea; A = πD²/4

A = (π x 0.018²)/4 = 2.5447 x 10^(-4) m²

Now, let's set up the equation.

-ρAv = (d/dt)(ρV) = V(dρ/dt) + ρ(dv/dt)

Thus,

Plugging in the relevant values to get;

-[1.225 x 2.5447 x 10^(-4) x 0.20] = 6 x 10^(-3)(dρ/dt) + (1.225 x -0.0001)

So,

-0.62345 x 10^(-4) = 6x10^(-3)(dρ/dt) - (1.225 x 10^(-4))

6x10^(-3)(dρ/dt) = (1.225 x 10^(-4)) - 0.62345 x 10^(-4)

6x10^(-3)(dρ/dt) = 0.60155 x 10^(-4)

(dρ/dt) = [0.60155 x 10^(-4)] /(6x10^(-3)) = 0.10026 x 10^(-1) = 0.010026 kg/m³.s

ρ

Answer:v=\sqrt{\frac{Gm}{r}}

Explanation:

Given

Orbital speed is v

Mass of planet is m

Radius of circular orbit is r

suppose M is the mas of satellite then centripetal force on satellite is equal to the Gravitational Pull.

\frac{Mv^2}{r}=\frac{GMm}{r^2}

where G=gravitational constant

thus on solving we get

v=\sqrt{\frac{Gm}{r}}

The orbital speed v of a satellite in a circular orbit of radius r around a planet of mass m should be v = √Gm/r.

Calculation of the orbital speed:

Since

Orbital speed is v

Mass of planet is m

The radius of the circular orbit is r

Here we assume M is the mass of the satellite so the centripetal force on the satellite should be equivalent to the Gravitational Pull.

So,

Mv^2 /r = GMm/r^2

Here

G=gravitational constant

So, v = √Gm/r.

Explanation:

• When the temperature of the water of the sanitizing solution is low, it means that there is more chance of "bacteria being present in the water".

• So this bacteria can go and affect the food and when the living being intake the food, there is a possibility for illness.

• How sanitizing hand is important, it is that equal important to sanitize the pans to "stay healthy".

• So, "the water should be poured out and new water containing the sanitizing solution should be used".

The food handler should not use the sanitizing solution to sanitize pans.

Who is a food handler?

A food handler simply means an individual that ensures that food is safe to eat and implements appropriate controls to minimize hazard.

In this case, the food handler should not use the sanitizing solution to sanitize pans. Rather, new sanitized water should be used.