<span>The combined
gas law has no official founder; it is simply the incorporation of the three
laws that was discovered. The combined gas law is a gas law that combines
Gay-Lussac’s Law, Boyle’s Law and Charle’s Law.
Boyle’s law states that pressure is inversely proportional with volume
at constant temperature. Charle’s law states that volume is directly
proportional with temperature at constant pressure. And Gay-Lussac’s law shows
that pressure is directly proportional with temperature at constant volume. The
combination of these laws known now as combined gas law gives the ratio between
the product of pressure-volume and the temperature of the system is constant.
Which gives PV/T=k(constant). When comparing a substance under different
conditions, the combined gas law becomes P1V1/T1 = P2V2/T2.</span>
Answer:
The Troposphere is the lowermost portion of Earth’s atmosphere. It is the densest layer of the atmosphere and contains approximately 75 percent of the mass of the atmosphere and almost all the water vapor and aerosol. The troposphere extends from the Earth’s surface up to the tropopause where the stratosphere begins.
Explanation:
Answer:
True. Diffusion and osmosis are forms of passive transport.
Explanation:
In diffusion, particles move from an area of higher concentration to one of lower concentration until equilibrium is reached.
In osmosis, a semipermeable membrane is present, so only the solvent molecules are free to move to equalize concentration.
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :



Work done on the wagon is given by the product of net force and displacement. It is given by :


W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity

m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting

The center of mass of the two-ball system is 7.05 m above ground.