A rectangular trough, 2.0 m long, 0.50 m wide, and 0.50 m deep, is completely full of water. One end of the trough has a small drain plug right at the bottom edge.
When you pull the plug, at what speed does water emerge from the hole? (in m/s)
Hint: V = sqrt (2 g H
I just don't know where to start.
the hint is the answer
v = sqrt (2*9.8*.5)
from
m g h = (1/2)mv^2
is
2 g h = v^2
the potential energy of a chunk of water at the surface is its kinetic energy exiting at the bottom
you can do it with Bernouli if you want to get sophisticated :)
p + rho g h + (1/2) rho v^2 = constant
at water surface p and v are 0
at bottom exit, p and h are 0
Thank you
To find the speed at which water emerges from the hole, you can use Bernoulli's principle, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline.
In this case, the water inside the trough is at rest before the plug is pulled. When the plug is opened, the water will flow out due to the force of gravity. Let's break down the problem step by step:
Step 1: Determine the height difference between the water level and the hole.
Since the water is completely full, the height difference (H) is the same as the depth of the trough, which is 0.50 m.
Step 2: Calculate the gravitational acceleration.
The gravitational acceleration (g) can be considered as approximately 9.8 m/s² on Earth.
Step 3: Use the formula to calculate the speed of water flow.
V = sqrt (2 g H)
Now, substitute the known values into the formula:
V = sqrt (2 * 9.8 * 0.50)
V = sqrt (9.8)
V = 3.13 m/s (rounded to two decimal places)
Therefore, the water will emerge from the hole at a speed of approximately 3.13 m/s.