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2 answers:
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Answer:
a) A = 0.603 m , b) a = 165.8 m / s² , c) F = 331.7 N
Explanation:
For this exercise we use the law of conservation of energy
Starting point before touching the spring
Em₀ = K = ½ m v²
End Point with fully compressed spring
=
= ½ k x²
Emo =
½ m v² = ½ k x²
x = √(m / k) v
x = √ (2.00 / 550) 10.0
x = 0.603 m
This is the maximum compression corresponding to the range of motion
A = 0.603 m
b) Let's write Newton's second law at the point of maximum compression
F = m a
k x = ma
a = k / m x
a = 550 / 2.00 0.603
a = 165.8 m / s²
With direction to the right (positive)
c) The value of the elastic force, let's calculate
F = k x
F = 550 0.603
F = 331.65 N
Answer : The equilibrium concentration of T(g) is 0.5 M
Solution :
Let us assume that the equilibrium reaction be:
The given equilibrium reaction is,
The expression of will be,
where,
= equilibrium constant = 16
[Z] = concentration of Z at equilibrium = 2.0 M
[R] = concentration of R at equilibrium = 2.0 M
[X] = concentration of X at equilibrium = 2.0 M
[T] = concentration of T at equilibrium = ?
Now put all the given values in the above expression, we get:
Therefore, the equilibrium concentration of T(g) is 0.5 M