we only see wavelengths from 400–700 nanometers.
Answer:
y = 80.2 mille
Explanation:
The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening
θ = 1.22 λ/ d
in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m
θ = 1.22 550 10⁻⁹ / 0.002
θ = 3.355 10⁻⁴ rad
Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi
tan θ = y / L
y = L tan θ
y = 2,389 10⁵ tan 3,355 10⁻⁴
y = 8.02 10¹ mi
y = 80.2 mille
This is the smallest size of an object seen directly by the eye
Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
![m_{1} v_{1i} + m_{2} v_{2i} + m_{3} v_{3i} = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}](https://tex.z-dn.net/?f=%20m_%7B1%7D%20%20v_%7B1i%7D%20%2B%20m_%7B2%7D%20%20v_%7B2i%7D%20%2B%20m_%7B3%7D%20%20v_%7B3i%7D%20%3D%20m_%7B1%7D%20%20v_%7B1f%7D%20%2B%20m_%7B2%7D%20%20v_%7B2f%7D%20%2B%20m_%7B3%7D%20%20v_%7B3f%7D%20)
Before the firecracker blows a coconut does not move, so left side is equal to 0:
![0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}](https://tex.z-dn.net/?f=%200%20%3D%20m_%7B1%7D%20v_%7B1f%7D%20%2B%20m_%7B2%7D%20v_%7B2f%7D%20%2B%20m_%7B3%7D%20v_%7B3f%7D%20)
We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
![v_{3f} = - \frac{m_{1} v_{1f} + m_{2} v_{2f} }{ m_{3} }](https://tex.z-dn.net/?f=v_%7B3f%7D%20%3D%20-%20%20%5Cfrac%7Bm_%7B1%7D%20%20v_%7B1f%7D%20%20%2B%20m_%7B2%7D%20v_%7B2f%7D%20%7D%7B%20m_%7B3%7D%20%7D%20)
We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
![v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s](https://tex.z-dn.net/?f=%20v_%7Bf%7D%20%3D%20%5Csqrt%7B%20v_%7B1f%7D%5E%7B2%7D%2B%20v_%7B2f%7D%5E%7B2%7D%20%7D%20%5C%5C%20%5C%5C%20v_%7Bf%7D%20%3D%2033.9m%2Fs)
This speed we can insert into formula for v3f:
![v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\ \\ v_{3f} = - \frac{2m*33.9 }{ 2m } \\ \\ v_{3f} = - 33.9m/s](https://tex.z-dn.net/?f=v_%7B3f%7D%20%3D%20-%20%5Cfrac%7Bm%2A33.9%2Bm%2A33.9%20%7D%7B%202m%20%7D%20%5C%5C%20%20%5C%5C%20v_%7B3f%7D%20%3D%20-%20%5Cfrac%7B2m%2A33.9%20%7D%7B%202m%20%7D%20%20%5C%5C%20%20%5C%5C%20v_%7B3f%7D%20%3D%20-%2033.9m%2Fs)
We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.