If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.
The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.
- A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m
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<span>The last option.
Plants absorb carbon dioxide from the atmosphere, water from the soil and other nutrients also from the soil - salts containing nitrogene, potassium, sulphur, etc. They use water and carbon dioxide to produce sugar through photosyntesis.
Decomposition is the reaction that converts any organic compound back into inorganic compounds - water, carbon dioxide and salts containing nitrogene, potassium, sulphur, etc. So it's basically the opposite.
So photosyntesis uses carbon dioxide as a reactive and take it from the atmosphere, whereas decomposition generates carbon dioxide as a product and releases it to the atmosphere.</span>
Answer:
15.065ft
Explanation:
To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.
By definition the drag force is expressed as:

Where
is the density of the flow
V = Velocity
= Drag coefficient
A = Area
For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3
For second Newton's Law the Force is also defined as,

Equating both equations we have:



Integrating


Here,






Replacing:




The flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.
<h3>How we compare the flow of electricity to water?</h3>
Water flowing in pipes is like flowing of electricity in a circuit. A battery is like a pump from where electricity comes and moves in the circuit. Electrons flowing through wires are like water molecules flowing through pipes. So in comparison between water and electricity, both water and electricity are similar to each other in flowing and movement.
So we can conclude that the flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.
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Answer:
Chloride and Salt
Explanation:
that make up table salt, a.k.a. sodium chloride (NaCl). ... As sodium chloride (NaCl) or calcium chloride (CaCl2) dissolve in water, ... the compound formed when a positive ion combines with a negative ion out of solution, but ... waters may have more if there is weathering or leaching from nearby mineral-rich soils and rocks.