Earths plates move, creating them to rub together to create earthquakes, and volcanoes occur because magma melts rocks and the pressure is so intense it causes the lava to leak out of the top of a mountain
Metal ores
Explanation:
in an area where subduction has occurred in times past, metal ores are likely to be found.
Metallic ores find subduction zone regions very favorable to crystallize out of a magma.
- Ores have different modes of formation.
- Typically, they are found in hydrothermal vents and black smokers of igneous intrusives.
- These are igneous terrains where metallic sulfides and other minerals crystallize out of magmatic body.
- Metals in magma usually have large sizes and do not partition easily in the melt.
At a subduction zone, partial melting of the subducting plate forces magma into nearby country rock as an intrusive and to the ocean floor where they form black smokers.
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components of the speed of the coin is given as




now the time taken by the coin to reach the plate is given by



now in order to find the height



so it is placed at 1.52 m height
Answer:
![F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]](https://tex.z-dn.net/?f=F_T%3D6k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bi%7D%2B10k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bj%7D%3D2k%5Cfrac%7BQ%5E2%7D%7BL%7D%5B3%5Chat%7Bi%7D%2B5%5Chat%7Bj%7D%5D)


Explanation:
I attached an image below with the scheme of the system:
The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:
![F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]](https://tex.z-dn.net/?f=F_T%3DF_Q%2BF_%7B3Q%7D%2BF_%7B4Q%7D%5C%5C%5C%5CF_T%3Dk%5Cfrac%7B%28Q%29%282Q%29%7D%7BR_1%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B%283Q%29%282Q%29%7D%7BR_2%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B%284Q%29%282Q%29%7D%7BR_3%7D%5Bcos%5Ctheta%20%5Chat%7Bi%7D%2Bsin%5Ctheta%20%5Chat%7Bj%7D%5D)
the distances R1, R2 and R3, for a square arrangement is:
R1 = L
R2 = L
R3 = (√2)L
θ = 45°
![F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]](https://tex.z-dn.net/?f=F_T%3Dk%5Cfrac%7B2Q%5E2%7D%7BL%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B6Q%5E2%7D%7BL%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B8Q%5E2%7D%7B%5Csqrt%7B2%7DL%7D%5Bcos%2845%5C%C2%B0%29%5Chat%7Bi%7D%2Bsin%2845%5C%C2%B0%29%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_T%3Dk%5Cfrac%7B2Q%5E2%7D%7BL%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B6Q%5E2%7D%7BL%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B8Q%5E2%7D%7B%5Csqrt%7B2%7DL%7D%5B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Chat%7Bi%7D%2B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_T%3D6k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bi%7D%2B10k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bj%7D%3D2k%5Cfrac%7BQ%5E2%7D%7BL%7D%5B3%5Chat%7Bi%7D%2B5%5Chat%7Bj%7D%5D)
and the magnitude is:

the direction is:

mass of iron block given as

density of iron block is

now the volume of the iron piece is given as


Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

here we know that
= density of liquid = 916 kg/m^3


Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium



So reading of spring balance will be 16.45 N
Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block



So the other scale will read 36.47 N