Answer:
Average acceleration on first part of the chunk is given as
![a_1 = 13.125 m/s^2](https://tex.z-dn.net/?f=a_1%20%3D%2013.125%20m%2Fs%5E2)
Average acceleration on second part of the chunk is given as
![a_2 = -13.125 m/s^2](https://tex.z-dn.net/?f=a_2%20%3D%20-13.125%20m%2Fs%5E2)
Explanation:
By momentum conservation along x direction we will have
![mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2](https://tex.z-dn.net/?f=mv_i%20%3D%20%5Cfrac%7Bm%7D%7B2%7Dv_1%20%2B%20%5Cfrac%7Bm%7D%7B2%7Dv_2)
so we have
![v_1 + v_2 = 2v](https://tex.z-dn.net/?f=v_1%20%2B%20v_2%20%3D%202v)
![v_1 + v_2 = 4.68](https://tex.z-dn.net/?f=v_1%20%2B%20v_2%20%3D%204.68)
also by energy conservation
![\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm%7D%7B2%7D%29v_1%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm%7D%7B2%7D%29v_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%2017%20J)
![\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7Dm%28v_1%5E2%20%2B%20v_2%5E2%29%20-%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%2017%20)
![(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)](https://tex.z-dn.net/?f=%28v_1%5E2%20%2B%20v_2%5E2%29%20-%202v%5E2%20%3D%20%5Cfrac%7B4%7D%7B7.7%7D%2817%29)
![(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83](https://tex.z-dn.net/?f=%284.68%20-%20v_2%29%5E2%20%2B%20v_2%5E2%20-%202v%5E2%20%3D%208.83)
![21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83](https://tex.z-dn.net/?f=21.9%20%2B%202v_2%5E2%20-%209.36%20v_2%20-%2010.95%20%3D%208.83)
![2v_2^2 - 9.36v_2 + 2.12 = 0](https://tex.z-dn.net/?f=2v_2%5E2%20-%209.36v_2%20%2B%202.12%20%3D%200)
by solving above equation we will have
![v_1 = 4.44 m/s](https://tex.z-dn.net/?f=v_1%20%3D%204.44%20m%2Fs)
![v_2 = 0.24 m/s](https://tex.z-dn.net/?f=v_2%20%3D%200.24%20m%2Fs)
Average acceleration on first part of the chunk is given as
![a_1 = \frac{4.44 - 2.34}{0.16}](https://tex.z-dn.net/?f=a_1%20%3D%20%5Cfrac%7B4.44%20-%202.34%7D%7B0.16%7D)
![a_1 = 13.125 m/s^2](https://tex.z-dn.net/?f=a_1%20%3D%2013.125%20m%2Fs%5E2)
Average acceleration on second part of the chunk is given as
![a_2 = \frac{0.24 - 2.34}{0.16}](https://tex.z-dn.net/?f=a_2%20%3D%20%5Cfrac%7B0.24%20-%202.34%7D%7B0.16%7D)
![a_2 = -13.125 m/s^2](https://tex.z-dn.net/?f=a_2%20%3D%20-13.125%20m%2Fs%5E2)
Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is located at 23.5°N and in December when the sub-solar point is at 23.5°S latitude.
Explanation:
At the sub-solar point, the sun strikes directly at the surface with an angle of 90 degrees at a given point.
Solistice refers to that point in time when the sun’s zenith is located at the farthest point from the equator.
During summer solistice on June 21, the sun’s zenith reaches northernmost point, sub-solar point is fixed at 23.5°S Tropic of Cancer making the earth tilt 23.4 degrees
During winter soliscitse on December 21, the sub-solar point is fixed at) Tropic of Capricorn.
Eh not really sure bout this one
The various contributions involved till the chapati is made is given below.
<h3>What is food?</h3>
The substance that we intake for the body to charge up by giving nutrients is called the food.
Wheat is a staple food. We make chapati from flour obtained from the wheat grains.
The various contributions involved till the chapati is made is given below.
Take required amount of atta in a container
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Add water accordingly to form a dough
↓
Apply oil to make dough smooth for long time
↓
Take small dough, make it a ball shaped and apply dry flour
↓
Roll it using rolling pin on the chapati maker plate
↓
After making it circular or any shape you want, place it on hot tawa
↓
Bake it on both the sides
↓
Chapati is ready
Thus, the flow chart is made.
Learn more about food.
brainly.com/question/16327379
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