What is the period of a blender blade that spins around 400 times in 5.0s
1 answer:
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Answer : The partial pressure of is, 67.009 atm
Solution : Given,
Partial pressure of at equilibrium = 30.6 atm
Partial pressure of at equilibrium = 13.9 atm
Equilibrium constant =
The given balanced equilibrium reaction is,
The expression of will be,
Now put all the values of partial pressure, we get
Therefore, the partial pressure of is, 67.009 atm
From the given information in the question, the correct option is Option 1: 14 cm.
A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:
PE = k
Where PE is the energy, k is the spring constant and x is extension.
i. Given that: PE = 10 J and x = 10 cm, then;
PE = k
10 = k
20 = 100k
k = 0.2 J/cm
ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.
PE = k
20 = (0.2)
40 = 0.2
= 200
x =
= 14.1421
x = 14.14 cm
So that;
x is approximately 14.00 cm.
Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.
For more information, check at: brainly.com/question/1352053.
Answer:
0.1 m
Explanation:
F = Force exerted on spring = 3 N
k = Spring constant = 60 N/m
x = Displacement of the block
As the energy of the system is conserved we have
The position of the block is 0.1 from the initial position.