Water enters a pipe with a cross sectional area of 3.0 cm^2 with a velocity of 3.0 m/s. The water encounters a constriction where its velocity is 15 m/s. What is the cross sectional area of the constricted portion of the pipe?
I figured it out.
To find the cross-sectional area of the constricted portion of the pipe, we can make use of the principle of continuity. According to this principle, the product of the cross-sectional area and the velocity of a fluid remains constant as it flows through a pipe with a constant volume.
Mathematically, this can be expressed as:
A1 * v1 = A2 * v2
where A1 and v1 are the cross-sectional area and velocity of the water before the constriction, and A2 and v2 are the cross-sectional area and velocity of the water after the constriction.
Given:
A1 = 3.0 cm² (convert it to m²: 1 cm² = 0.0001 m², so A1 = 3.0 * 0.0001 m² = 0.0003 m²)
v1 = 3.0 m/s
v2 = 15 m/s
We can rearrange the equation above to solve for A2:
A2 = (A1 * v1) / v2
Substituting the given values:
A2 = (0.0003 m² * 3.0 m/s) / 15 m/s
A2 = 0.00006 m²
Therefore, the cross-sectional area of the constricted portion of the pipe is 0.00006 square meters.