Water is running out of a faucet, falling straight down, with an initial speed of 0.50 m/s. At what distance below the faucet is the radius of the stream reduced to one-half its value at the faucet?
To find the distance below the faucet where the radius of the water stream is reduced to one-half its initial value, we need to apply the principle of conservation of mass for an incompressible fluid.
Let's denote the radius of the water stream at the faucet as R₀ and the distance below the faucet as x, where the radius reduces to R₀/2. Assuming that the water stream is cylindrical, the volume of water flowing per second at the faucet (Q₀) is equal to the volume of water flowing per second at a distance x below the faucet (Qₓ).
According to the principle of conservation of mass, the volume of water flowing per second is constant along the stream. Therefore, we can express this principle mathematically as:
π * R₀² * v₀ = π * (R₀/2)² * vₓ
Where:
π represents the mathematical constant pi (approximately 3.14159)
R₀ is the initial radius of the water stream at the faucet
v₀ is the initial speed of the water stream
vₓ is the speed of the water stream at a distance x below the faucet
We are given that the initial speed of the water stream is 0.50 m/s. Substituting these values into the equation, we get:
π * R₀² * 0.50 = π * (R₀/2)² * vₓ
The π's cancel out, and simplifying the equation further, we have:
R₀² * 0.50 = (R₀/2)² * vₓ
Now, let's solve for vₓ:
vₓ = (R₀² * 0.50) / (R₀/2)²
vₓ = 2 * (R₀² * 0.50) / R₀²
vₓ = 2 * 0.50
vₓ = 1.00 m/s
Therefore, the speed of the water stream at a distance x below the faucet is 1.00 m/s.
To find the distance x, we can use equations of motion. We need to find the time it takes for the speed to change from 0.50 m/s to 1.00 m/s, and then calculate the distance traveled during that time.
Using the equation of motion:
v = u + at
Where:
v is the final velocity (1.00 m/s)
u is the initial velocity (0.50 m/s)
a is the acceleration (assumed to be constant at 9.8 m/s² due to gravity)
t is the time it takes to achieve the final velocity
We can rearrange the equation to solve for t:
t = (v - u) / a
t = (1.00 - 0.50) / 9.8
t = 0.50 / 9.8
t ≈ 0.051 seconds
To find the distance x, we can use another equation of motion:
s = ut + 0.5at²
Where:
s is the distance traveled
u is the initial velocity (0.50 m/s)
t is the time (0.051 seconds)
a is the acceleration (9.8 m/s²)
Substituting the given values into the equation, we have:
s = 0.50 * 0.051 + 0.5 * 9.8 * 0.051²
s ≈ 0.025 m
Therefore, the distance below the faucet where the radius of the water stream is reduced to one-half its initial value is approximately 0.025 meters.
That will happen when the velocity of the water has increased by a factor of 4.
(stream Area)*(velocity) = constant.
Use conservation of energy to get that distance.