If, as is typical, each of them breathes about 500 cm3 of air with each breath, approximately what volume of air (in cubic meter
1 answer:
Answer:
<em>a) 12614.4 m^3</em>
<em>b) 28.8 m</em>
Explanation:
The complete question is
Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm^3 of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?
The average breathing rate is 12 breaths per minute
there are 60 minutes x 24 hours x 365 days in a year = 525600 minutes in a year
if an average human takes 12 breath per minute, then in a year an average human take 12 x 525600 = 6307200 breath
For the four astronauts, the amount of breath = 4 x 6307200 = 25228800 breath in total.
The volume of air per breath = 500 cm^3
1 cm^3 = 10^-6 m^3
500 cm^3 = m^3
= 500 x 10^-6 = 5 x 10^-4 m^3
Therefore in a year, the volume of these astronauts breath in a year = 5 x 10^-4 x 25228800 = <em>12614.4 m^3</em>
b) If the space station is spherical, the volume will be given as =
where r is the radius of the space station
This volume of the space station must be able to contain all the volume of breath produced by the astronauts which is = 12614.4 m^3
Equating, we have
12614.4 =
12614.4 =
12614.4 = 4.1893
= 12614.4/4.1893 = 3011.1
r = =<em> 14.4 m</em>
<em>diameter of the space station = 14.4 m x 2 = 28.8 m</em>
You might be interested in
Answer:
A) v = 28.3 m/s
B) t = 4.64 s
Explanation:
A)
- Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:
- Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:
- So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:
- Δh = 26.0 m + 14. 8 m = 40.8 m (3)
- Replacing now in (1), we can solve for vf, as follows:
B)
- In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
- 1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
- 2) Time elapsed from this point until it hits the street, with vo=0.
- For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:
- Replacing by the givens in (5) and solving for Δt, we get:
- For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:
- Replacing by the givens and solving for t in (7), we get:
- So, total time is just the sum of (6) and (8):
- t = 2.9 s + 1.74 s = 4.64 s