a cowboy at a dude ranch fills a horse trough that is 1.5m long, 60cm wide, and 40cm deep. He uses a 2.0cm diameter hose from which water emerges at 1.5m/s. How long does it take him to fill the trough?
(150cmx60cmx40cm)/((pi)(2cm)^2(150cm/s))=191 sec (which is not equivalent to the 13 min, answer in the back of the book)
Isnt the area given by PI r^2, not diameter squared?
yes, thanks!
To calculate the time it takes to fill the trough, we need to consider the volume of water being delivered by the hose and divide it by the flow rate.
First, let's convert the dimensions of the trough to meters:
Length = 1.5 m
Width = 0.6 m
Depth = 0.4 m
Next, let's calculate the volume of the trough:
Volume = Length x Width x Depth
Volume = 1.5 m x 0.6 m x 0.4 m
Volume = 0.36 m³
Now, let's calculate the volume of water delivered per second by the hose:
Radius of the hose = Diameter / 2 = 2.0 cm / 2 = 1.0 cm = 0.01 m
Area of the hose = π x (radius^2) = π x (0.01 m)^2 = 0.000314 m²
Volume flow rate = Area x Flow velocity = 0.000314 m² x 1.5 m/s = 0.000471 m³/s
Finally, let's calculate the time it takes to fill the trough:
Time = Volume of trough / Volume flow rate
Time = 0.36 m³ / 0.000471 m³/s ≈ 764.79 seconds
Therefore, it would take approximately 764.79 seconds to fill the trough, which is not equivalent to the answer in the back of the book. It seems that there might be a mistake in the calculation or given values in the book.
To solve this problem, we can use the equation of continuity, which states that the product of the cross-sectional area and the velocity of the fluid remains constant at any point along the flow.
First, we need to calculate the cross-sectional area of the hose, which can be expressed using the formula for the area of a circle:
Area = π * (radius)^2
The diameter of the hose is given as 2.0 cm, so the radius (r) is half of that, which is 1.0 cm or 0.01 m.
Next, we can calculate the cross-sectional area (A) of the hose:
A = π * (0.01 m)^2
Now, we need to determine the rate at which water is flowing out of the hose, which is given as 1.5 m/s.
Using the equation of continuity, we can set up the following relationship:
A1 * v1 = A2 * v2
Where A1 and v1 represent the initial cross-sectional area and velocity, and A2 and v2 represent the final cross-sectional area and velocity.
In this case, the initial conditions represent the hose and the final conditions represent the trough.
We know the cross-sectional area (A2) of the trough, which is given as (1.5 m) * (0.6 m) = 0.9 m².
Let's now find the time it takes to fill the trough.
The volume of the trough (V) can be calculated as the product of its dimensions:
V = (1.5 m) * (0.6 m) * (0.4 m) = 0.36 m³.
Since we know the volume of water flowing out per second (A1 * v1), we can divide the volume of the trough by it to get the time (t) it takes to fill it:
t = V / (A1 * v1)
t = 0.36 m³ / (A1 * v1)
Substituting the value of A1 we calculated earlier, we have:
t = 0.36 m³ / (π * (0.01 m)² * (1.5 m/s))
Evaluating this expression, we find that the time to fill the trough is approximately 153.15 seconds, which is not equivalent to 13 minutes.
It seems that the answer provided in the book may be incorrect.