Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
![W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7BL%7D%20Fdx%20%3D%20%5Cint_%7B0%7D%5E%7BL%7Dax%20%5C%2C%20dx%20%3D%20%5Cfrac%7Ba%7D%7B2%7D%20%20%5Bx%5E%7B2%7D%20%5D_%7B0%7D%5E%7BL%7D%20%3D%20%20%5Cfrac%7BaL%5E%7B2%7D%7D%7B2%7D%20)
Answer:
Answer:
toward the center
Explanation:
Before answering, let's remind the first two Newton Laws:
1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force
2) An object acted upon a net force F experiences an acceleration a according to the equation

where m is the mass of the object.
In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).
So, the correct answer is
toward the center
We are in the lowest layer called Troposphere.
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Answer:
1keff=1k1+1k2
see further explanation
Explanation:for clarification
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?
From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.
Now the spring is in series combination
F
e
F=ke
k=f/e.........*
where k is the force constant or the constant of proportionality
k=f/e
............................1
also for effective force constant
divide all through by extension
1) Total force is
Ft=F1+F2
Ft=k1e1+k2e2
F = k(e1+e2) 2)
Since force on the 2 springs is the same, so
k1e1=k2e2
e1=F/k1 and e2=F/k2,
and e1+e2=F/keq
Substituting e1 and e2, you get
1/keq=1/k1+1/k2
Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.