Given that,
A system dissipates 12 J of heat into the surroundings.
28 J of work is done on the system.
To find,
The internal energy of the system.
Solution,
The first law of thermodynamics is used here. According to this law,
Q is heat and W is work done
Here,
Q = -12 J is the heat dissipated by the system
W = -28 J is the work done on the system
ATQ,
So, the change of internal energy of the system is 16 J.
The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:
where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get
Therefore. the elastic potential energy stored in the spring is 6.75J .
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Answer:
F' = 64 F
Explanation:
The electric force between charges is given by :
Where
q₁ and q₂ are charges
r is the distance between charges
When each charge is doubled and the distance between them is 1/4 its original magnitude such that,
q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)
New force,
Apply new values,
So, the new force becomes 64 times the initial force.