Water flows at a rate of 8000 cubic inches per minute into a cylindrical tank. the tank has a diameter of 128 inches and a height of 72 inches. what is the height, in inches, of the water in the tank after 10 minutes? round your answer to the nearest tenth.

The height of the water = 6.2 in. to the nearest tenth

Step-by-step explanation:

∵ The rate of flows of water into the tank = 8000 in.³/min.

∴ The volume of the water in the tank after 10 min. = 8000 × 10 = 80000 in.³

∵ The water take the shape of the tank

∴ The height of the water = volume of water ÷ Area of the base of the tank

∵ The tank is cylinder with diameter 128 in. and height 72 in.

∴ The area of the base of the tank = π(128/2)²

∴ The height of the water =

∴ The height of the water = 6.2 in. to the nearest tenth

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this