Hi there!
Recall the equation for centripetal force;
m = mass (kg)
v = velocity (m/s)
r = radius (m)
We are not given the mass directly, we are given the object's weight.
Now, we can plug this value along with the given velocity and radius to solve:
What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?
1 answer:
The elastic potential energy of the spring is 6.8 J
Explanation:
The elastic potential energy of a compressed/stretched spring is given by the equation:
where
k is the spring constant
x is the elongation of the spring
The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):
By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,
Therefore when the spring has a elongation of , its potential energy is
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Answer:
Explanation:
Halogen group:This group contains Cl,F,Br and I
There are 7 electrons in the outermost shell of each element of halogen family.
Each element of this family required one electron to fill their complete outermost shell.
We know that
Cl=17=
Therefore, the general configuration for atoms of the halogen group is given by
When the ball on a spring is replaced by a smaller ball with half the mass : The period decrease by 29%
<h3>Calculate the Period </h3>
We will apply the formula below to calculate period
T =
where :
T₁ / T₂ = where m₂ = m₁ / 2
Therefore :
T₂ = = 0.71 T
Hence we can conclude that the period will decrease by 29% ( i.e 100 - 71 ).
Learn more about Period calculation : brainly.com/question/10728818