The distance between the two charges is 
Explanation:
The electrostatic force between two charged objects is given by Coulomb's law:

where:
is the Coulomb's constant
are the charges of the two objects
r is the separation between the two charges
In this problem, we are given the following:



Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

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This would be an example of deposition (Phase transition).
<span>a) 13 seconds
b) 130 m/s
The formula for the distance an object moves while under constant acceleration is d = 1/2AT^2. So let's define d as 830 m, A as 9.8m/s^2, and solve for T
830 m = 1/2 9.8 m/s^2 T^2
830 m = 4.9 m/s^2 T^2
Divide both sides by 4.9 m/s^2
169.3878 s^2 = T^2
Take the square root of both sides
13.01491 s = T
Since we only have 2 significant figures, round the result to 13 seconds which is the answer to the first part of the question. To find out how fast the marble is moving, just multiply T and A together
13 s * 9.8 m/s^2 = 127.4 m/s
Since we only have 2 significant figures, round the result to 130 m/s.</span>
To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.
By definition we know that the change in entropy is given by

Where,
Q = Heat transfer
T = Temperature
On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

According to the data given we have to,




PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is



On the other hand,



The total change of entropy would be,



Since
the heat engine is not reversible.
PART B)
Work done by heat engine is given by



Therefore the work in the system is 100000Btu
Answer:
d
Explanation:
because is a reducing agent, O 2 is an oxidizing agent.