To find the average of data collected, add all of the measurements: 5.9+6.2+6.3+6= 12.2
Then, divide the total amount by the number of data collected which is 4: 12.2/4= 3.05
The average speed of the runner of the race is approximately 3.05 km/min
Feel free to ask me any other questions you might have :)
Answer:
Live training (whether through videoconference or a live course on a learning management system) or an eLearning course with knowledge checks is suggested so that learners can receive immediate feedback. Inclusion of a post-test – as well as an electronic guide describing jurisdiction-specific protocols – is strongly recommended.
Explanation:
Explanation:
The coordinates of points are:
A (0, 70, 0),
B (40, 0, 0), C (40, 0, 40) D (60, 0, 60).
The position vectors for corresponding cables are:
the unit vectors for these positions are:
![u_{AD} =r_{AD} /[r_{AD} ]=(-60/110)i-(70/110)j-(60/110)k\\=-0.5455i-0.6364j-0.5455k](https://tex.z-dn.net/?f=u_%7BAD%7D%20%3Dr_%7BAD%7D%20%2F%5Br_%7BAD%7D%20%5D%3D%28-60%2F110%29i-%2870%2F110%29j-%2860%2F110%29k%5C%5C%3D-0.5455i-0.6364j-0.5455k)
![u_{AC} =r_{AC} /[r_{AC} ]=-(40/90)i-(70/90)j+(40/90)k\\=-0.444i-0.778j+0.444k](https://tex.z-dn.net/?f=u_%7BAC%7D%20%3Dr_%7BAC%7D%20%2F%5Br_%7BAC%7D%20%5D%3D-%2840%2F90%29i-%2870%2F90%29j%2B%2840%2F90%29k%5C%5C%3D-0.444i-0.778j%2B0.444k)
![u_{AB} =r_{AB}/[ r_{AB} ]=(40/80.6)j-(70/80.6)j+0k\\=0.4963i-0.8685j+0k](https://tex.z-dn.net/?f=u_%7BAB%7D%20%3Dr_%7BAB%7D%2F%5B%20r_%7BAB%7D%20%5D%3D%2840%2F80.6%29j-%2870%2F80.6%29j%2B0k%5C%5C%3D0.4963i-0.8685j%2B0k)
The factors are:
![F_{AB} =[F_{AB} ]u_{AB} =0.9926i-1.737j+0k\\\\F_{AC}= [F_{AC]} u_{AC} =-0.8888i-1.5556j+0.8888k](https://tex.z-dn.net/?f=F_%7BAB%7D%20%3D%5BF_%7BAB%7D%20%5Du_%7BAB%7D%20%3D0.9926i-1.737j%2B0k%5C%5C%5C%5CF_%7BAC%7D%3D%20%5BF_%7BAC%5D%7D%20u_%7BAC%7D%20%3D-0.8888i-1.5556j%2B0.8888k)
![F_{AD} =[F_{AD}] u_{AD} =-1.0910i-1.2728j-1.0910k](https://tex.z-dn.net/?f=F_%7BAD%7D%20%3D%5BF_%7BAD%7D%5D%20u_%7BAD%7D%20%3D-1.0910i-1.2728j-1.0910k)
so The resultant force exerted on tower by cables are:
![F_{R} =F_{AB}+ F_{AC}+ F_{AD} \\=-0.9875i-4.5648j-0.2020k\\magnitude=[F_{R} ]=4.674kN](https://tex.z-dn.net/?f=F_%7BR%7D%20%3DF_%7BAB%7D%2B%20F_%7BAC%7D%2B%20F_%7BAD%7D%20%5C%5C%3D-0.9875i-4.5648j-0.2020k%5C%5Cmagnitude%3D%5BF_%7BR%7D%20%5D%3D4.674kN)
