Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.

This question is incomplete, the complete question is;

The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters

a) Find the volume V in terms of x and L

b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.  

a) the volume V in terms of x and L is  ((√3/4)x²L) m³

b) required expression is (2/(3)^(1/u))√(At/L)

Explanation:

Given that;

form the question and image below;

triangular prism ends are equilateral triangle

side length = x meter

Dimension of the prism = L meter

Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter

Volume of the triangular prism = Area × height

= √3/4 (x)² × L

V = ((√3/4)x²L) m³

Therefore, the volume V in terms of x and L is  ((√3/4)x²L) m³

b)

Rate of water entering = A m³/hr

Depth of water tank = d meter

Time = t

Length of prism = L

now Rate of water entering is A m³/hr

dv/d = A                             [  V = ((√3/4)x²L) m³ ]

and

dv/dt = √3/4 [2x dx/dt ] L                   { L is constant }

so

A = √3/4 [2x dx/dt ] L  

∫A dt = √3/2 [ Lx dx ]                   { Integrate both sides}

At = √3/2 × Lx × x²/2

x² = uAt / √3L                              { we find square root of both sides}

x = √( uAt / √3L )

x = (2/(3)^(1/u))√(At/L)

Therefore; required expression is (2/(3)^(1/u))√(At/L)

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.