EDKP 395 Lecture Notes - Lecture 16: Lactic Acid, California State Route 2, Ac3D

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limereindeer57

20 Nov 2016

School

McGill University

Department

Kinesiology and Physical Ed

Course

EDKP 395

Professor

Dennis Jensen

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Document Summary

Breathing pattern: minute ventilation v"e & breathing pattern. Tidal volume vt & frequency fr. Ventilation v"e: alveolar ventilation v"a & breathing pattern. Same minute ventilation v"e & dead space ventilation v"d. From rest to exercise = want deep & slow pattern. So body volume 1st & frequency later. *frequency fr will minimize alveolar ventilation v"a so maximize w/ tidal volume vt. Eventually = increase too much tidal volume vt so elastic recoil & wob work of breathing. Increase more tidal volume vt than frequency fr. Diffusion of gases = maintly determined by driving pressure gradient to prevent respiratory muscle stealing from exercising muscle. Total pressure of gas mixture = equal to sum of pressures that each gas would exert. Independently sum of partial pressures of individual gases in mixture. Partial pressure = % concentration of specific gas total pressure of gas mixture: percent & partial pressures of gases in dry ambient air - sea level.

Related Questions

A space habitat for a long space voyage consists of two cabins eachconnected by a cable to a central hub as shown in the figure below.The cabins are set spinning around the hub axis, which is connectedto the rest of the spacecraft to generate artificial gravity.
(a) What forces are acting on an astronaut in one of thecabins?

(b) Write Newton's second law for an astronaut lying on the "floor"of one of the habitats, relating the astronaut's mass m, hisvelocity v, his radial distance from the hub r, and the normalforce n. (Use any variable or symbol stated above asnecessary.)

(c) What would n have to equal if the 49.4-kg astronaut is toexperience half his normal Earth weight?

(d) Calculate the necessary tangential speed of the habitat fromNewton's second law.
tangential speed=?

(e) Calculate the angular speed from the tangential speed.

(f) Calculate the period of rotation from the angular speed.

(g) If the astronaut stands up, will his head be moving faster,slower, or at the same speed as his feet? Why?

(i) Calculate the tangential speed at the top of his head if he is2.08 m tall.

1. If a car's wheels are replaced with wheels of greater diameter,will the reading of the speedometer change? Explain.

2. Weightlessness in space is known to have harmful effects on thebodies of people in space. To set up inhabitants (people) in space,it has been suggested that rotating cylinders about 10 miles longand 5 miles in diameter be placed in space. The purpose of theirrotation is to simulate gravity for the people so they have weight.Explain how this is possible?
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Please use DragMath to submit the solutions to problems No. 5 and 6of Activity 9A on page 129 of Bendall & Howison.
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1. Why do the string instruments go flat (lower frequencies) andthe wind instruments go sharp (higher frequencies) during aperformance if an orchestra doesn't warm up beforehand?
2. If you were carrying a fish tank filled with water, what willhappen if your walking frequency matches the natural frequency ofthe water in the tank?
Good luck with these questions. I look forward to a heartydiscussion by Friday this week.
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1. Household circuits are always wired so that the electricaldevices are connected in parallel. Why is this so?
2. How are circuit breakers connected in the household circuits(series or parallel)? What is the use of a circuit breaker? Why isit more advantageous to use a circuit breaker compared to using afuse?
Hope to hear your answers soon.
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Mathematical Question:
Try the question on capacitors Activity 14C no. 3. You can useDragmath to submit your solution. If you are scanning your answersand uploading a file, please keep its size small.
Challenge Question:
A fully charged parallel-plate capacitor remains connected to abattery while you slide a dielectric between the plates. Do thefollowing quantities increase, decrease, or stay the same? (a)capacitance, C (b) charge, Q (c) electric field, E between theplates, (d) change in voltage, V (e) energy stored in thecapacitor.
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Have a go; looking forward to some interesting replies.
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Metal detectors in airports: When you walk through the doorway ofan airport metal detector, you are really walking through a coil ofmany turns, i.e. an inductor connected to a resonant circuit tunedto the resonant frequency when there is no metal in the inductor.So how does this circuit detect any metal that is in your pocketwhen you walk through the metal detector?
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A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

2 Answers

Relevance

electron1
Lv 7

4 years ago

There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Let’s use the following equation to determine the acceleration.

vf2 = vi^2 + 2 * a * d

6.5^2 = 1.4^2 + 2 * a * 12.4

24.8 * a = 40.29

a = 40.29 ÷ 24.8

This is approximately 1.62 m/s^2.

Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

This is approximately 82 N.

Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

Net force = F + (1187.76 ÷ 12.4) – 41

Set the two net forces equal to each other.

F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

F is approximately 27.3 N

Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J
JupiterSky
Lv 6

4 years ago

His change in kinetic energy is due to gravitational- and his own work,

while the friction does negative work.

∆K = ∆U +∆W + ∆Wf

½m(Vf²−Vi²) = mgh + ∆W + Ff*d

∆W = ½m(Vf²−Vi²) − mgh − Ff*d

∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

∆W = 508 J

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Next Question

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asked in Science & MathematicsPhysics · 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the
spring (located on the left-hand side of the
track) and compresses the spring a distance
4.7 cm from its equilibrium position (as shown
in the ﬁgure below). The block starts from
rest, is accelerated by the compressed spring,
and slides across a frictionless horizontal track
(as shown in the ﬁgure below). It leaves the
track horizontally, ﬂies through the air, and
subsequently strikes the ground.
acceleration = 9.81
 
A) What is the spring constant?
B)What is the speed v of the block when it leaves
the track?
C) What is the total speed of the block when it
hits the ground?
...Show more

Update:

Height Above ground = 1.9 m
Lands = 4.59m away

What is the spring constant? please help me!!!!?

3 Answers

Relevance

Jánošík
Lv 7

8 years ago

Favorite Answer

I'm sure there is some crucial information in "the figure below" !!!

Edit:

OK... now we got something to work with ... ;-)

Time needed to fall a vertical distance of 1.9 m is:

t = √( 2h / g )

t = √[ 2×(1.9 m) / (9.81 m/s²) ]

t = 0.6224 s

In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

r = v×t

(4.59 m) = v×(0.6224 s)

v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

That is also the speed of the block as it leaves the spring.

Its kinetic energy at that time is:

KE = m×v²/2

KE = (0.487 kg)×(7.3749 m/s)²/2

KE = 13.244 J

That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

KE = PEe = k×d²/2

(13.244 J) = k×(0.047 m)²/2

k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

The gravitational potential energy of the block on the table is:

PEg = m×g×h

PEg = (0.487 kg)×(9.81 m/s²)×(1.9)

PEg = 9.077 J

The total energy of the block on the table is:

TE = PEg + PEe

TE = (9.077 J) + (13.244 J)

TE = 22.32 J

When hitting the ground, all that energy is converted to kinetic energy, so

TE = KE = m×v²/2

(22.32 J) = (0.487 kg)×v²/2

v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

...Show more
Technobuff
Lv 7

8 years ago

There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??
STALKER
Lv 4

8 years ago

use f=kl

where k is the spring constant and l is extension from the natural length.

k=f/l

since you have mass in grams. use w=mg

w=0.487 x 9.8 = 4.77N

Now use the equation,

k=4.77/4.7

k=1.012

Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

...Show more

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Next Question

ひいらぎ

Lv 5

asked in Science & MathematicsPhysics · 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.
 
A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.
-What is the moment of Inertia of the wash load?
-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2
...Show more

Physics Momentum, Rolling?

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

I can't visualize the first problem.

The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

...Show more

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Next Question

asked in Science & MathematicsPhysics · 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.
 
(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree
 
(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s
...Show more

Physics:You and your friends push a 75-kg...Help!?

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

♣the distance the pig was pushed up is

s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

♠ kinetic energy of the pig at the moment it’s released after push-up is

E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

♦pig’s potential energy at vertex (when stopped) is E1=mgh;

the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

f=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;

f= μ*mg*cos(b);

♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;

(μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;

μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence

(a); b=atan(μ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5°;

♥ pig’s potential energy above the low end of the ramp is

W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);

Pig’s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

and W1=0.5*m*v^2; thus

0.5*m*v^2 = mg*(s*sin(b) +h) – μ*mg*cos(b)*(s + h/sin(b));

v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;

(b); v =1.605 m/s;

...Show more

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EDKP 395 Lecture Notes - Lecture 16: Lactic Acid, California State Route 2, Ac3D

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