Answer:
A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.
B) The Diameter of this spherical space station should be 30.4m
Explanation:
The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:
If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.
The volume of a sphere is:
So the diameter is:
![D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m](https://tex.z-dn.net/?f=D%3D2r%3D2%5Csqrt%5B3%5D%7B%5Cfrac%7B3V_%7Bsph%7D%7D%7B4%5Cpi%7D%7D%20%3D30.4m)
Answer:
True
Explanation:
A crowbar makes our work easier by multiply effort because it belongs to first class lever.
And first class lever makes work easier by multiplying the effort
The answer is 60 mph.
The speed (v) is distance (d) per time (t): v = d/t
Car A:
v1 = ?
t1 = 2 h
d1 = ?
___
v1 = d1/t1
d1 = v1 * t1
Car B:
v2 = ?
t2 = 1.5 h
d2 = ?
___
v2 = d2/t2
d2 = v2 * t2
<span>Two cars traveled equal distances:
d1 = d2
</span>v1 * t1 = v2 * t2
<span>Car B traveled 15 mph faster than Car A:
v2 = v1 + 15
</span>v1 * t1 = v2 * t2
v2 = v1 + 15
________
v1 * 2 = (v1 + 15) * 1.5
2v1 = 1.5v1 + 22.5
2v1 - 1.5v1 = 22.5
0.5v1 = 22.5
v1 = 22.5/0.5
v1 = 45 mph
v2 = v1 + 15
v2 = 45 + 15
v2 = 60 mph
Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage
.
The balance point is not on the potentiometer wire - this statement means that
. In that case ,
l > L
V > E
Answer:
ΔU = 5.21 × 10^(10) J
Explanation:
We are given;
Mass of object; m = 1040 kg
To solve this, we will use the formula for potential energy which is;
U = -GMm/r
But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.
Thus;
ΔU = -GMm((1/r_f) - (1/r_i))
Where;
M is mass of earth = 5.98 × 10^(24) kg
r_f is final radius
r_i is initial radius
G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Since, it's moving to altitude four times the Earth's radius, it means that;
r_i = R_e
r_f = R_e + 4R_e = 5R_e
Where R_e is radius of earth = 6371 × 10³ m
Thus;
ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)
× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))
ΔU = 5.21 × 10^(10) J
