A) lighting an electric lamp as it becomes darker
We make a graphic of this problem to define the angle.
The angle we can calculate through triangle relation, that is,
![sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%20%5Cfrac%7Bc%7D%7BQP%7D%5C%5Csin%5Ctheta%20%3D%20%5Cfrac%7Bc%7D%7BR%7D%5C%5C%5Ctheta%3Dsin%5E%7B-1%7D%5Cfrac%7Bc%7D%7BR%7D)
With this function we should only calculate the derivate in function of c
![\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7Bc%5E2%7D%7BR%5E2%7D%7D%7D%28%5Cfrac%7Bc%7D%7BR%7D%29%27%5C%5C%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7BR%5E2-c%5E2%7D%7D)
That is the rate of change of
.
b) At this point we need only make a substitution of 0 for c in the equation previously found.
![\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%5Cbig%7C_%7Bc%3D0%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7BR%5E2-0%7D%7D%5C%5C%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%5Cbig%7C_%7Bc%3D0%7D%20%3D%20%5Cfrac%7B1%7D%7BR%7D)
Hence we have finally the rate of change when c=0.
Answer:
156.96 N
Explanation:
F=ma where m is the mass and a is acceleration
Substituting 16 Kg for m and 9.81 m/s2 for g then
F=16*9.81= 156.96 N
M = 4kg
F =8N
a..?
F =m.a
8 = 4.a
a = 2m/s^2