1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Elis [28]
3 years ago
13

Why do heavier people have a lower blood and alcohol level?

Physics
1 answer:
Ulleksa [173]3 years ago
4 0
In general, the less you weigh the more you will be affected by a given amount of alcohol. Alcohol has a high affinity for water. Basically, one's blood alcohol concentration is a function of the total amount of alcohol in one's system divided by total body water. So for two individuals with similar body compositions and different weights, the larger individual will achieve lower alcohol concentrations than the smaller one if ingesting the same amount of alcohol.
<span>However, for people of the same weight, a well-muscled individual will be less affected than someone with a higher percentage of fat since fatty tissue does not contain very much water and will not absorb very much alcohol.</span>
You might be interested in
I need homework help
KIM [24]
1) the weight of an object at Earth's surface is given by F=mg, where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is 
F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N

2) On Mars, the value of the gravitational acceleration is different:g=3.7 m/s^2. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 
g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg: 
g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as 
g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2

<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
</span>F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N<span>
</span>
5 0
3 years ago
A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo
gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

                       H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

 Height at any time 'T' after the drop =

                          (initial height) +

                                              (initial velocity) x (T) +
                                                                 (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

     Upward  = the positive direction

                       Initial height = +150 m
                       Initial velocity = + 3 m/s

     Downward = the negative direction

                     Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

                 H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
0    =  (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

                   -4.9 T²  +  3T  + 150  =  0

Use the quadratic equation:

                         T  =  (-1/9.8) [  -3 plus or minus √(9 + 2940)  ]

                             =  (-1/9.8) [  -3  plus or minus  54.305  ]

                             =  (-1/9.8) [ 51.305  or  -57.305 ]

                          T  =  -5.235 seconds    or    5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

                    H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
H'(t)  =  v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set                                  v₀ + a T  =  0

                                      +3  -  9.8 T  =  0

Add 9.8 to each  side:   3               =  9.8 T

Divide each side by  9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh !  I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
                       (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) =  1.5 m/s .

Distance climbed during that time = (average speed) x (time)

                                                           =  (1.5 m/s) x (0.306 sec)

                                                           =  0.459 meter  (hardly any at all)

     But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at  150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely  5.847 seconds after its release.  

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work.  The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0
3 years ago
Could I please get some help on this question I don’t understand .
Oksana_A [137]

Answer:

12.5 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 8 m

Final velocity (v) at 8 m above the lowest point =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

The velocity of the roller coaster at 8 m above the lowest point can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 8)

v² = 0 + 156.8

v² = 156.8

Take the square root of both side

v = √156.8

v = 12.5 m/s

Therefore, the velocity of the roller coaster at 8 m above the lowest point is 12.5 m/s.

5 0
3 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
3 years ago
In the deep ocean, a water wave with wavelength 95 m travels at 12 m/s. Suppose a small boat is at the crest of this wave, 1.2 m
dimaraw [331]

Answer:

the vertical position is 1.1971m

Explanation:

Recall that

y = 1.2 * cos (\frac{2*\pi * distance travelled}{wavelenght})

recall that v = \frac{distance}{time}

thus;  distance =  v* t

this implies that distance = 12 * 5.0

                                          =  60

therefore;  y = 1.2 cos \frac{2*3.142*60}{95}

                  y = 1.1971m

the

3 0
3 years ago
Other questions:
  • 1. A goal is scored in soccer when_____. (1 point)
    10·2 answers
  • Which of the following statements are true for magnetic force acting on a current-carrying wire in a uniform magnetic field? Che
    14·1 answer
  • What is the difference between a negative feedback system and a positive feedback system?
    7·1 answer
  • Which of the following statements is NOT true?
    13·1 answer
  • How fast would the car need to go to double its kinetic energy?
    13·2 answers
  • Use this free body diagram to help you find the magnitude of the force F1 needed to keep this block in static equilibrium 15.3 N
    6·1 answer
  • Who are people that you can go to for help name three?
    9·2 answers
  • Which feature of an object affects its weight? Select three options.
    12·1 answer
  • What type of energy does friction produce?
    6·1 answer
  • A wave has a wavelength of 3 m and a frequency of 340 Hz. How far will the wave go in 20 s?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!