a water hose is used by a gardener a to fill a 30-liter bucket {one liter =1 000cm^3}. nozzele with an opening of cross sectional area0.500cm^2 is then attached to the hose.the nozzlemis held so that water is projected horizontally from a point 1.00m above the ground. over what horizontal distance can the water be projected?
find veloicty at the nozzle
velocity=volumerate/area
you have no volume rate...ie, 30 liter bucket in WHAT time?
To find the horizontal distance the water can be projected, we need to use the principles of projectile motion. Let's break down the problem step-by-step.
Step 1: Convert the volume of water in the bucket to cubic centimeters (cm³).
30 liters x 1000 cm³/liter = 30,000 cm³
Step 2: Calculate the velocity of water leaving the nozzle.
The velocity of water leaving the nozzle can be found using Torricelli's law: v = sqrt(2gh), where:
- v is the velocity,
- g is the acceleration due to gravity (9.8 m/s²), and
- h is the height (1.00 m)
Let's convert the height to meters (m):
1.00 m = 100 cm
Substituting these values into the formula:
v = sqrt(2 * 9.8 m/s² * 100 cm)
v = sqrt(1960)
v ≈ 44.27 m/s
Step 3: Calculate the horizontal distance traveled by the water.
In the absence of air resistance, the horizontal distance (d) can be calculated using the equation: d = v * t, where:
- d is the distance,
- v is the horizontal component of velocity, and
- t is the time of flight.
Since the water is projected horizontally, the horizontal component of velocity is equal to the velocity of the water leaving the nozzle. The initial vertical velocity is zero.
Let's calculate the time of flight (t):
Using the equation h = 0.5gt², we can find t:
1.00 m = 0.5 * 9.8 m/s² * t²
2.00 = 4.9t²
t² ≈ 0.41
t ≈ sqrt(0.41)
t ≈ 0.64 s
Now we can calculate the horizontal distance:
d = v * t
d ≈ 44.27 m/s * 0.64 s
d ≈ 28.3 m
Therefore, the water can be projected horizontally over a distance of approximately 28.3 meters.
To solve this problem, we need to use the principles of fluid mechanics and projectile motion.
First, let's calculate the volume of water passing through the nozzle per unit time. We know that the cross-sectional area of the nozzle is 0.500 cm^2, which is equal to 0.500 cm^2 * (1 m^2 / 10,000 cm^2) = 0.0000500 m^2.
The gardener is filling a 30-liter (or 30,000 cm^3) bucket. We can convert this volume to cubic meters by dividing it by 1,000,000 (since 1 liter = 1,000 cm^3). So the volume of water passing through the nozzle per second is 30,000 cm^3 / 1,000,000 = 0.0300 m^3.
Next, we need to determine the velocity of the water exiting the nozzle. Since the water is projected horizontally, there is no vertical component to the velocity. We can use the principle of conservation of energy to find the velocity. The water initially has gravitational potential energy, which is converted into kinetic energy as it exits the nozzle:
mgh = (1/2)mv^2
Where:
m = mass of water
g = acceleration due to gravity
h = height from which the water is projected
v = velocity of the water
The mass of water can be calculated as follows:
density of water (ρ) = mass per unit volume
density of water = 1000 kg/m^3 (approximately)
Volume of water passing through the nozzle per unit time = 0.0300 m^3
So, the mass of water passing through the nozzle per second = density of water * volume of water = 1000 kg/m^3 * 0.0300 m^3 = 30 kg/s.
Now, plugging this value along with the acceleration due to gravity g = 9.81 m/s^2 into the equation, we can solve for the velocity (v):
(30 kg/s) * (9.81 m/s^2) * h = (1/2) * (30 kg/s) * v^2
Simplifying this equation, we find:
v = sqrt(2 * 9.81 * h)
Substituting the given value for h = 1.00 m into the equation, we find:
v = sqrt(2 * 9.81 * 1.00) = 4.43 m/s
Now that we have the velocity, we need to determine the horizontal distance traveled by the water. Since there is no horizontal acceleration acting on the water, it will travel horizontally at a constant velocity (v).
We can use the following equation of motion to calculate the horizontal distance (d):
d = v * t
We need to determine the time (t) it takes for the water to hit the ground. To do this, we can use the equation of motion for vertical motion:
h = (1/2) * g * t^2
Since the water is projected horizontally from a height of 1.00 m, we can set h = 1.00 m and solve for t:
1.00 m = (1/2) * (9.81 m/s^2) * t^2
Simplifying this equation, we find:
t = sqrt((2 * 1.00 m) / (9.81 m/s^2)) = 0.451 s
Finally, we can substitute the values of v = 4.43 m/s and t = 0.451 s into the formula for distance (d):
d = (4.43 m/s) * (0.451 s) = 1.99 m
Therefore, the water can be projected over a horizontal distance of approximately 1.99 meters.