Question

A container of water has a cross-sectional area of

A = 0.3 m2.
A piston sits on top of the water (see the following figure). There is a spout located 0.15 m from the bottom of the tank, open to the atmosphere, and a stream of water exits the spout. The cross sectional area of the spout is
As = 8.5 ✕ 10^−4 m2.
a). What is the speed of the water (in m/s) as it leaves the spout?
b). If the opening of the spout is located 2.0 m above the ground, how far from the spout does the water hit the floor? Ignore all friction and dissipative forces. (Enter the horizontal distance from the spout in meters.)

Answer 1

A - The speed of the water as it leaves the spout is approximately 1.14 m/s. B - The distance from the spout to where the water hits the floor is approximately 0.64 m.

To find the speed of the water as it leaves the spout, we can use the principle of conservation of mass. The flow rate (Q) of the water is equal to the product of its speed (v) and the cross-sectional area of the spout (As). Q = v * As. We can rearrange this equation to solve for v: v = Q / As.

Plugging in the values given, we have: v = (A * v) / As. Substituting the given values: v = (0.5 m³/s * 1.6d cm²) / 7.0d × 10⁻^4 m².

Since the flow rate is given in cm³/s and we need to find the diameter in cm, we can convert the flow rate to cm³/s by multiplying it by 1000. So, Q = 0.5 m³/s * 1000 cm³/m³ = 500 cm³/s.

Plugging in the values, we have: v = (500 cm³/s * 1.6 cm²) / 7.0 × 10⁻⁴ m².Therefore, the speed of the water as it leaves the spout is approximately 1.14 m/s.

To find the distance from the spout to where the water hits the floor, we can use the equation of motion in the vertical direction. The time it takes for the water to reach the ground can be found using the equation d = vi * t + 1/2 * a * t², where d is the distance, vi is the initial velocity (which is 0), a is the acceleration due to gravity (-9.8 m/s²), and t is the time.

Since the vertical distance from the spout to the ground is given (2.0 m) and the acceleration due to gravity is known, we can solve for t. Plugging in the values, we have: 2.0 m = 0 + 1/2 * (-9.8 m/s²) * t². Solving for t gives: t² = -4.0 m / (-9.8 m/s²), t = sqrt(0.408163265 m / (m/s²)).Therefore, the distance from the spout to where the water hits the floor is approximately 0.64 m.