Answer:
Though the process photosynthesis
Explanation:
One of the important ways by which the atmosphere recycle gases through interactions with the the biosphere is by photosynthesis.
The biosphere is the living component on the earth. The atmosphere is the gaseous envelope round the earth.
- Photosynthesis is the process whereby green plants manufacture their food using carbon dioxide and water in the presence of sunlight to produce glucose and oxygen gas.
- This way, atmospheric carbon dioxide is exchanged for oxygen gas.
This way, there is a sustained interaction between the atmosphere and biosphere.
Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:

Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:

Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
Complete Question
Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?
Answer:
Go-cart A is faster
Explanation:
From the question we are told that
The length of the track is 
The speed of A is 
The uniform acceleration of B is 
Generally the time taken by go-cart A is mathematically represented as
=> 
=> 
Generally from kinematic equation we can evaluate the time taken by go-cart B as

given that go-cart B starts from rest u = 0 m/s
So

=>
=>
Comparing
we see that
is smaller so go-cart A is faster
Nice couch lol
and aluminum I think
the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period
Thank you