Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000Pa. At the nozzle the pressure decreases to atmospheric pressure (101300 Pa), thereis no change in height. Use the Bernoulli equation to calculate the velocity of thewater exiting the nozzle. (Hint: The density of water is 1000 kg/m3 and gravity g is9.8 m/s2. Pay attention to units!)]a. 14m/sb.12 m/sc.7m/sd.13 m/s
Question
Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000Pa. At the nozzle the pressure decreases to atmospheric pressure (101300 Pa), thereis no change in height. Use the Bernoulli equation to calculate the velocity of thewater exiting the nozzle. (Hint: The density of water is 1000 kg/m3 and gravity g is9.8 m/s2. Pay attention to units!)]a. 14m/sb.12 m/sc.7m/sd.13 m/s
Solution
To solve this problem, we can use the Bernoulli equation, which relates the pressure, velocity, and height of a fluid.
The Bernoulli equation is given by:
P + 1/2 * ρ * v^2 + ρ * g * h = constant
Where: P is the pressure ρ is the density of the fluid v is the velocity of the fluid g is the acceleration due to gravity h is the height of the fluid
In this case, we are given the following information: Initial velocity (v1) = 1.0 m/s Initial pressure (P1) = 200000 Pa Final pressure (P2) = 101300 Pa Density of water (ρ) = 1000 kg/m^3 Gravity (g) = 9.8 m/s^2
Since there is no change in height, the height term can be ignored.
Using the Bernoulli equation, we can set up the following equation:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Substituting the given values:
200000 + 1/2 * 1000 * (1.0)^2 = 101300 + 1/2 * 1000 * v2^2
Simplifying the equation:
200000 + 500 = 101300 + 500 * v2^2
200500 = 101300 + 500 * v2^2
Subtracting 101300 from both sides:
99200 = 500 * v2^2
Dividing both sides by 500:
v2^2 = 198.4
Taking the square root of both sides:
v2 = √198.4
v2 ≈ 14.07 m/s
Therefore, the velocity of the water exiting the nozzle is approximately 14 m/s.
So, the correct answer is option a. 14 m/s.
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