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Brut [27]
2 years ago
7

A student is pedaling a stationary bicycle at the U of M Rec Center. Her alveolar PO2 = 115 mm Hg and her oxygen consumption is

1.0 liter/min. If the inspired PO2 = 150 mm Hg and the barometric pressure is 747 mm Hg, what is her alveolar ventilation?
Physics
1 answer:
polet [3.4K]2 years ago
8 0

Answer:

This is an alveolar gas equation question and it is used to approximate the partial pressure of oxygen in the alveolus (PAO2):

The equation states;

PₐO₂ = P₁O₂ - (PₐCO₂/R) = [(PB - PH₂O) × F₁O₂ - (PₐCO₂/R)]   ...................Eqn 1

where

PₐO₂ = Alveolar partial pressure of O2 = 115mmHg

P₁O₂ = Inspired partial pressure of O2 = 150mmHg

PB = barometric pressure,

PH₂O = Water vapor pressure (usually 747 mmHg),

F₁O₂ = fractional concentration of inspired oxygen,

and R = gas exchange ratio. (Usually around 0.8)

FₐO₂ = Fraction of alveolar O₂

(O₂) = 1L/min = 1dm³

From eqn 1. we have

PₐCO₂ = (P₁O₂ - PₐO₂)/R

   = (150 - 115)x0.8

PₐCO₂ = 28mmHg

Similarly from Eqn 1, we have

F₁O₂ = (PₐO₂ + PₐCO₂/R)/(PB - PH₂O)

F₁O₂ = (115 + (28/.8))/(747 - 47)

F₁O₂ = 0.21

Now to find the Alveolar Ventilation A, we will use this equation;

O₂ = A(F₁0₂ - FₐO₂)                                                                   .................Eqn 2

But FₐO₂ = PₐO₂/(PₐO₂ + PₐCO₂)

FₐO₂ = 115/ (115+28) = 0.8

A = O₂/(F₁0₂ - FₐO₂)

A = 0.001/(0.21 - 0.8)

A = 0.00169m³/min

Hence, the aveolar ventilation is 0.00169m³/min

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Sloan [31]

Answer:

Explanation:

Time to cover first 100 km = 1 hour.

time remaining = 3.15 - 1 = 2.15 hour .

Time to cover next 42 km = 1 hour .

Time remaining = 2.15-1 = 1.15 hour.

Distance to be covered = 310 - 142

= 168 km

least speed needed = distance remaining / time remaining

= 168 / 1.15

= 146.08 km / h .

4 0
3 years ago
A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.
Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0
2 years ago
Force of 10N down,10N to the right,and 5N to the left are acting on a ball .it acceleration horizontally to the right.what other
denpristay [2]

Answer:

A 10 N force pointing up

Explanation:

If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).

Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

Then there must also be a 10 N force pointing up acting on the object.

3 0
3 years ago
Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis
Veronika [31]

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

3 0
3 years ago
Understanding the benefits of an activity can __________.
Bogdan [553]

Answer:A.

Increase your motivation to continue doing it

Explanation:

benefits help you

8 0
3 years ago
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