A fire truck attaches two fire hoses to a hydrant to fight a fire. Each hose has a
diameter of 20 cm. The underground water pipe going to the hydrant has a diameter of 100 cm. In
the underground pipe, the water has a speed of 2.0 m/s.
the combined hoses have an area 2/25 that of the pipe.
So, the water in the hoses travels at a speed 25/2 * 2 = 25 m/s
Well, that fire truck sure knows how to make a splash! With two fire hoses attached to a hydrant, each with a diameter of 20 cm, they're ready to bring the heat down. But let's not forget about the underground water pipe, which has a diameter of 100 cm and water rushing through it at a speedy 2.0 m/s.
Now, the first thing I have to say is that the underground water pipe must be feeling a bit jealous. You see, with a diameter of 100 cm, it's quite the tank. Meanwhile, those fire hoses with their 20 cm diameter are like the little brothers trying to keep up.
But when it comes to speed, that's where the water in the underground pipe shines. With a velocity of 2.0 m/s, it's like the Usain Bolt of water pipes. It's all about that speed, baby!
So, even though those fire hoses may not match up in size, they can still pack a punch. And with the water rushing through the underground pipe, it's a comedy in motion. Just imagine the water coming out of those hoses, doing its best to extinguish the fire, while the water from the underground pipe is like a supporting actor, giving a helping hand.
In the end, it's a teamwork of water forces, coming together to fight the fire. So, let's give them all a round of applause!
To answer your question step-by-step:
Step 1: Calculate the cross-sectional area of each hose.
Given that the diameter of each hose is 20 cm, the radius can be calculated as (20/2) = 10 cm = 0.1 m.
The cross-sectional area of each hose can be calculated using the formula: A = π * r^2
Substituting the radius (0.1 m) into the formula, we get: A = 3.14 * (0.1)^2 ≈ 0.0314 m^2
Step 2: Calculate the cross-sectional area of the underground water pipe.
Given that the diameter of the water pipe is 100 cm, the radius can be calculated as (100/2) = 50 cm = 0.5 m.
The cross-sectional area of the water pipe can be calculated using the formula: A = π * r^2
Substituting the radius (0.5 m) into the formula, we get: A = 3.14 * (0.5)^2 ≈ 0.785 m^2
Step 3: Calculate the speed of water in the fire hoses.
Since the volume flow rate is constant, we can use the equation: A1 * v1 = A2 * v2, where A1 and v1 are the cross-sectional area and speed of water in the underground pipe, and A2 and v2 are the cross-sectional area and speed of water in the hoses.
Using the given values, we can rearrange the equation to solve for v2: v2 = (A1 * v1) / A2
Substituting the values, we get: v2 = (0.785 * 2.0) / 0.0314 ≈ 50 m/s
Step 4: Analyze the result.
The speed of water in the fire hoses is approximately 50 m/s. This high speed is necessary to ensure adequate pressure and water flow to fight the fire effectively.
To find the speed of water coming out of each fire hose, we can apply the principles of conservation of mass and continuity equation.
The continuity equation states that the mass flow rate of a fluid is constant at any given point in an incompressible flow. Mathematically, it can be expressed as:
A1V1 = A2V2,
where A1 and A2 are the cross-sectional areas of the pipe and hose respectively, and V1 and V2 are the velocities of the water flowing through them.
Given:
Diameter of the underground pipe (D1) = 100 cm = 1 meter
Diameter of each fire hose (D2) = 20 cm = 0.2 meters
Velocity of water in the underground pipe (V1) = 2.0 m/s
We can first find the cross-sectional areas of the pipe and the hoses using their diameters:
Area of the underground pipe (A1) = (π/4)(D1)^2
= (3.14/4)(1^2)
= 0.7854 square meters
Area of each fire hose (A2) = (π/4)(D2)^2
= (3.14/4)(0.2^2)
= 0.0314 square meters
Now, let's substitute the values into the continuity equation:
A1V1 = A2V2
0.7854 * 2.0 = 0.0314 * V2
1.5708 = 0.0314 * V2
V2 = 1.5708 / 0.0314
V2 ≈ 50 m/s
Therefore, the speed of water coming out of each fire hose is approximately 50 m/s.