Joseph remembers reading that for a particular pressure, P1, there is a maximum height to which a fluid can rise. He reasons that in this case the pressure (P2) must drop to zero so that the water can no longer flow to increased heights. He decides to check if the simulation can be made to demonstrate such conditions. He decides to make A2 as small as possible in the simulation (2.2 cm2) so that he increases the speed of the fluid in tube 2 as high as is possible (he decides to do this after examining the continuity equation). He then calculates the height of the fluid under these conditions where the pressure would fall to zero so that the fluid could no longer be made to flow under the conditions set up in the simulation. He then runs the simulation under the largest Δy allowed to check his calculation. What does Joseph calculate for the maximum height?

h = P₁ / 9800

Explanation:

This is a fluid mechanics problem, let's write the Bernoulli equation at two points, the subscript 1 for the lowest point and the subscript of 2 for the point with the highest height.

           P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

at the highest point P₂ = 0 and v₂ = 0,

          P1 + ½ ρ v12 = ρ g (y₂ -y₁)

we use the continuity equation for the velocity at the lowest point

          A₁ v₁ = A₂ v₂

Since the velocity at the highest point is zero, this implies from the equation that the velocity at the lowest point is also zero. In the no-flow condition

             P₁ = ρ g (y₂ -y₁)

             h = y₂-y₁

             h = P₁ /ρ g

the density of water is ρ = 1000 kg / m³ and g = 9.8 m/s², we substitute

              h = P₁ / 9800

Let's do a calculation, suppose that P₁ = 1 10⁵ Pa

             h = 1 10⁵ / 9800

             h = 10.2 m

it is easily magnetized.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)