Experience research study. experiment explain scientific process useful for research for an action . Albert enistein . He was found all the scientific knowledge. more scientist same method
Having said that,
mV + M0 = (m + M)v; where m = bullet mass, M = block mass, V = bullet velocity upon incoming, and v = the block and bullet velocity (ideal without losses) is the answer you are looking for. Thus, v = V(m/(m + M)) is what you want. m = Mb, M = Mw, and V = Vi.
Honed I don’t know where the question is
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)
Answer:
water height: 3cm
oil height: 12 cm
Explanation:
70 cm = 0.7 m
Suppose both arms have the same cross-section area S:
Let water density be
Let x be the height of water in the 2nd arm, then 4x is the heigh of oil in the 2nd arm.
For the system to be balanced, then the fluid mass of both arms must be the same
Where V1 is the fluid volume is the 1st arm, Vw is the water volume in the 2nd arm, Vo is the oil volume in the 2nd arm,
is the fluid height in the 1st arm,
is the oil density.
We can divide both sides by S
So the water height in the 2nd arm is 3 cm and the oil height is 3*4 = 12 cm