the branch of optics that studies interference, diffraction, polarization, and other phenomena for which the ray approximation of geometric optics is not valid.
Answer:
T=7.4 N hence T<30 N
Explanation:
The figure is likely to be similar to the one attached. Writing the equation for forces we have
F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as
30 N - T = (30/9.81)a
Also, we know that T=F*a/g and substituting 10N for F we obtain the second equation as
T = (10/9.81)a
Adding the first and second equations we obtain
30 = 4.077471967
a Hence
![a=\frac {30}{4.077471967}=7.3575 m/s^{2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%20%7B30%7D%7B4.077471967%7D%3D7.3575%20m%2Fs%5E%7B2%7D)
and T=a hence
T is approximately 7.4 N
ANSWER
![\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%2A%7D%204.67%2A10%5E%7B-9%7D%5Ctext%7B%20%7DN%20%5Cend%7Bequation%2A%7D)
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
![F=\frac{GMm}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D)
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
![\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20F%3D%5Cfrac%7B6.67430%2A10%5E%7B-11%7D%2A70%2A1%7D%7B1%5E2%7D%20%5C%5C%20%20%5C%5C%20F%3D4.67%2A10%5E%7B-9%7D%5Ctext%7B%20%7DN%20%5Cend%7Bgathered%7D)
That is the answer.
Because of the pole and the generator you would have to biuld