Answer:
Rapid human population growth has a number of consequences. Population grows fastest in the world's poorest countries. High fertility rates have historically had a strong relation with poverty, and high childhood mortality rates
Explanation:
i got it from goo.gle but i hope that helps:)
have a good day bro
Suppose you have a cylinder filled with diatomic oxygen (O2) and it is running low. The cylinder is shown above, is made of steel, and has a fixed volume of 10 L.
You are asked to determine the number of O2 molecules that are left in the cylinder, so you take a measurement of the temperature to be 20℃. You then note that the pressure gauge reads 100 psi, which you checked at sea level in Bellingham, where the local pressure is one atm (14.7 psi). Calculate the number of O2 molecules left in the container.
Answer:
The number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.
Explanation:
The number of oxygen molecules can be found using the Ideal Gas law:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 100 psi
V: is the volume = 10 L
n: is the number of moles =?
T: is the temperature = 20 °C = 293 K
R: is the gas constant = 0.082 L*atm/(K*mol)
Hence, the number of moles is:
[tex]n = \frac{PV}{RT} = \frac{100 psi*\frac{1 atm}{14.7 psi}*10 L}{0.082 L*atm/(K*mol)*293 K} = 2.83 moles[/tex]
Now, the number of molecules can be found with Avogadro's number:
[tex]n_{m} = \frac{6.022 \cdot 10^{23}\: molecules}{1\: mol}*2.83 moles = 1.70 \cdot 10^{24} \: molecules[/tex]
Therefore, the number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.
I hope it helps you!
Which factors affect the gravitational force between two objects?
-
distance and velocity
O mass and distance
O mass and weight
acceleration and weight
TELE
Answer:
mass and distance
Explanation:
mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.
An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion, away from the space station. The astronaut has a mass of ma=102 kgma=102 kg and the bag of tools has a mass of mb=10.0 kg.mb=10.0 kg. If the astronaut is moving away from the space station at vi=2.10 m/svi=2.10 m/s initially, what is the minimum final speed vb,fvb,f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?
Answer:
The answer is "[tex]2.352 \ \frac{m}{s}[/tex]"
Explanation:
[tex]\to mass(m_1)=102 \ kg\\\\\to mass(m_2)=10 \ kg \\\\\to v=2.10\ \frac{m}{s}\\\\[/tex]
momentum before:
[tex]\to p=(m_1+m_2)v[/tex]
[tex]=(102+10)2.10\\\\=(102\times 2.10 +10 \times 2.10)\\\\=214.2+21\\\\=235.2[/tex]
momentum After:
[tex]\to p=(m_1+m_2)v[/tex]
[tex]=(102\times 0 +10 \times v)\\\\ =(0 +10v)\\\\=10v\\[/tex]
Calculating the conservation of momentum:
[tex]\to \text{momentum before = momentum After}[/tex]
[tex]\to 235.2=10v\\\\\to v= \frac{235.2}{10}\\\\ \to v=2.352 \ \frac{m}{s}[/tex]