The diffusion coefficient of the gas is proportional to the average rate of thermal motion of the molecules.
the average velocity is inversely proportional to the square root of the molar mass
so
The gas diffusion rate is inversely proportional to the square root of its molecular weight.
I would say Option B) because Option C) is wrong since matter cannot be created. A closed system does not exchange matter so it's not Option D). Since an island is an isolated area, Option A) is wrong.
Answer:
Increasing its charge
Increasing the field strength
Explanation:
For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:
![qvB = m\frac{v^2}{r}](https://tex.z-dn.net/?f=qvB%20%3D%20m%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
q is the charge
v is the velocity
B is the magnetic field
m is the mass
r is the radius of the orbit
The period of the motion is
![T=\frac{2\pi r}{v}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B2%5Cpi%20r%7D%7Bv%7D)
Re-arranging for r
![r=\frac{Tv}{2\pi}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7BTv%7D%7B2%5Cpi%7D)
And substituting into the previous equation
![qvB = m \frac{Tv^3}{2\pi}](https://tex.z-dn.net/?f=qvB%20%3D%20m%20%5Cfrac%7BTv%5E3%7D%7B2%5Cpi%7D)
Solving for T,
![T=\frac{2\pi q B}{m v^2}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B2%5Cpi%20q%20B%7D%7Bm%20v%5E2%7D)
So we see that the period is:
- proportional to the charge and the magnetic field
- inversely proportional to the mass and the square of the speed
So the following will increase the period of the particle's motion:
Increasing its charge
Increasing the field strength
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
1 Joule IS 1 newton-meter.