A roller coaster is stationary at the top of a ramp.

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

polet [3.4K]

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

8 0

2 years ago

How do you work out Potential Difference??? <br> Please can you make it simple :) <br> Thanks.

patriot [66]

Answer:

Potential difference is the work done in moving a positive test charge from infinity to the point in question.

Voltage is an expression of PD. (Joules / Coulomb)

Say that a capacitor has a PD of 5 Volts. The work in moving a positive test charge from the positive plate to the negative plate is -5 Joules/Coulomb or -5 volt. (At the positive plate the positive test charge (1 Coulomb) already has a PD of + 5 Volts.)

7 0

2 years ago

**URGENT, I WILL PAY 30 POINTS, PLEASE HELP**

melamori03 [73]

All three windows are the same size.

A has 10 complete waves visible through the window. B has 3, and C has 4.

So A must have the smallest wavelengths.

4 0

3 years ago

Read 2 more answers

A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s

Sveta_85 [38]

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg = sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute 2.955 * 10 ^6 J for W and 120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

3 0

2 years ago

Ultrasonic images are obtained from the inside organs of our body. This process uses which property of sound wave?

Temka [501]

This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.

The process of ultrasonic images uses the "echo" property of the sound waves.

Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

Learn more about echo here:

brainly.com/question/14335186?referrer=searchResults

The attached picture shows the process of ultrasonic imaging.

4 0

1 year ago