Answer:
A low difference in the concentration of the molecule across the media
Explanation:
Diffusion is a type of passive transport where the molecules move in the influence of concentration gradient of diffusing molecules i.e. from the higher concentration region to the lower concentration region. There are some factors which affect the rate of diffusion, these are written below -
- Mass of diffusing molecule - lighter molecules diffuse faster and heavier one diffuse relatively slower.
- Concentration gradient - rate of diffusion is higher if the difference in concentration of the diffusing particles is larger in the two regions.
- Distance traveled - molecules diffuse faster if they need to travel little distance during diffusion.
- Temperature - rate of diffusion will be greater at higher temperatures because the movement of diffusing molecules gets increased.
- Solvent density - rate of diffusion tend to be lower if the solvent has higher density.
Looking at these factors we can conclude that the second statement in the question tells about a negative impact regarding the diffusion because due to low difference in concentration across the two media, the rate of diffusion will be lower.
The angular momentum is defined as,

Acording to this text we know for conservation of angular momentum that

Where
is initial momentum
is the final momentum
How there is a difference between the stick mass and the bug mass, we define that
Mass of the bug= m
Mass of the stick=10m
At the point 0 we have that,

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity
vector from the point of reference (O), and ve is the velocity
At the end with the collition we have

Substituting




Applying conservative energy equation we have


Replacing the values and solving

Substituting
l=\frac{13}{0.54(9.8)}

Answer: The correct explanation is 2.
Explanation: The warm air is less dense (it expands) and thus it is lighter than the cold air so it will rise up to the floor. Therefore, when you place the heater on the floor it will warm the cold air which would then rise and be replaced by more cold air which would again get warm and rise and so on until the room is heated. This means that the correct explanation is 2.
On the other hand, if you put the heater at the ceiling, it will warm the cold air near the ceiling which would stay up there (it is lighter than the cold air under it). This means that the only way for the heat to spread from this ceiling level warm air to the lower levels is via conduction which is slow.
I think you forgot to give the choices along with the question. I am answering the question based on my research and knowledge. <span>If a layer was deposited but does not appear in the rock record, the thing that happened is erosion. I hope that this is the ans wer that has actually come to your desired help.</span>