Answer:
The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.
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Answer:
4km
Explanation:
15 minutes is 1/4 of an hour.
1/4 of 16 is 4.
the theory and development of computer systems able to perform tasks that normally require human intelligence, such as visual perception, speech recognition, decision-making, and translation between languages.
To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:
![I = I_0 cos^2\theta](https://tex.z-dn.net/?f=I%20%3D%20I_0%20cos%5E2%5Ctheta)
Where,
Indicates the intensity of the light before passing through the polarizer
I = Resulting intensity
= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light
From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore
![I_1= \frac{I_0}{2}](https://tex.z-dn.net/?f=I_1%3D%20%5Cfrac%7BI_0%7D%7B2%7D)
In the case of the second polarizer the angle is directly 60 degrees therefore
![I_2 = I_1 cos^2\theta](https://tex.z-dn.net/?f=I_2%20%3D%20I_1%20cos%5E2%5Ctheta)
![I_2 = (\frac{I_0}{2} ) cos^2(60)](https://tex.z-dn.net/?f=I_2%20%3D%20%28%5Cfrac%7BI_0%7D%7B2%7D%20%29%20cos%5E2%2860%29)
![I_2 = 0.125I_0](https://tex.z-dn.net/?f=I_2%20%3D%200.125I_0)
In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be
![\theta_3 = 90-60 = 30](https://tex.z-dn.net/?f=%5Ctheta_3%20%3D%2090-60%20%3D%2030)
Then,
![I_3 = I_2 cos^2\theta_3](https://tex.z-dn.net/?f=I_3%20%3D%20I_2%20cos%5E2%5Ctheta_3)
![I_3 = 0.125I_0 cos^2 (30)](https://tex.z-dn.net/?f=I_3%20%3D%200.125I_0%20cos%5E2%20%2830%29)
![I_3 = 0.09375I_0](https://tex.z-dn.net/?f=I_3%20%3D%200.09375I_0)
Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.