The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.
<h3>Work done by the applied force</h3>
The area under force versus displacement graph is work done.
The total work done by pulling the object 7 m, can be grouped into two areas;
- First area, A1 = area of triangle from 0 m to 2.0 m
- Second area, A2 = area of trapezium, from 2.0 m to 7.0 m
A1 = ¹/₂ bh
A1 = ¹/₂ x (2) x (20)
A1 = 20 J
A2 = ¹/₂(large base + small base) x height
A2 = ¹/₂[(7 - 2) + (7-3)] x 50
A2 = ¹/₂(5 + 4) x 50
A2 = 225 J
<h3>Total work done </h3>
W = A1 + A2
W = 20 J + 225 J
W = 245 J
Learn more about work done here: brainly.com/question/8119756
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.
Answer:
If I’m correct 300 joules
Explanation:
Answer:
2.5 m/s
Explanation:
There are calculators online that can help you easily calculate the accerlation.