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:
D.None of these
Explanation:
The derivation of acceleration formula:
Let us call the 5kg mass 
and the 4kg mass 
. If the tension in the string is 
then for the mass
(1). 
<em>(the negative sign on the right side indicates that acceleration is downwards)</em>
And for the mass
(2). 
<em> (the acceleration is upwards, hence the positive sign)</em>
Solving for in the 2nd equation we get:
,
and putting this into the 1st equation we get:
Back to the question:
Using the formula for the acceleration we find
which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.
Yes that is a balaned equation
Its C because if it is a low frequency it will not change much so it will be a longer wavelength and the higher the frequency the shorter the wavelength
Answer:
film is at distance of 3.07 cm from lens
Explanation:
Given data
focal length = 3.06 cm
distance = 10.4 m = 1040 cm
to find out
How far must the lens
solution
we apply here lens formula that is
1/f = 1/p + 1/q
here f = 3.06 and p = 1040 so we find q
1/f = 1/p + 1/q
1/3.06 = 1/1040 + 1/q
1/ q = 0.3258
q = 3.0690 cm
so film is at distance of 3.07 cm from lens
