A large container, 114 cm deep is filled with water. If a small hole is punched in its side 93.0 cm from the top, at what initial speed will the water flow from the hole?
SAAAME
Mass of water that escapes per unit time
= m
Since the mass of water drops in height by an equivalent of h=1.14 m, we can equate change in energy = 0,
or
mgh-(1/2)mv^2=0
Solve for v
v=(2hg)^1/2
v=(2*93*980)^1/2
v=426.9cm=4.269m
To determine the initial speed at which water will flow from the hole, we can use Torricelli's law, which relates the velocity of the fluid flow from an opening to the height of the fluid above the opening.
Torricelli's Law states that the velocity, v, of an ideal fluid flowing from an orifice is given by the equation:
v = sqrt(2gh)
Where:
v = velocity of the fluid flow
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the water surface above the opening
First, let's convert the depth and distance from centimeters to meters for consistency:
Depth of the container, h = 114 cm = 1.14 m
Distance of the hole from the top, h_hole = 93.0 cm = 0.93 m
Next, we need to determine the height of the water surface above the hole. We can calculate this height by subtracting the distance of the hole from the total depth of the container:
h_water = h - h_hole
= 1.14 m - 0.93 m
= 0.21 m
Now, substitute the values into the equation to find the velocity:
v = sqrt(2 * 9.8 m/s^2 * 0.21 m)
= sqrt(4.116 m^2/s^2)
≈ 2.03 m/s
Therefore, the initial speed at which the water will flow from the hole is approximately 2.03 m/s.