A gardener disregards the current hosepipe ban and water restrictions and uses a
hosepipe to fill a 30.0 L bucket. They notice that it takes 1.00 min to fill the bucket. A
nozzle with an opening of cross-sectional area 0.500 cm2
is then attached to the hose.
The nozzle is held so that water is projected horizontally from a point 1.00 m above the
ground. Over what horizontal distance is the water projected?
Tutor initi
To find the horizontal distance over which the water is projected, we need to find the initial velocity of the water jet.
First, let's calculate the volume of water that flows through the hosepipe in one second:
Volume of water = 30.0 L = 30.0 dm³
There are 1000 cm³ in 1 dm³, so the volume of water in cm³ is:
Volume of water = 30.0 dm³ × 1000 cm³/dm³ = 30000 cm³
Since it takes 1 minute to fill the bucket, the volume of water flowing through the hosepipe in 1 second is:
Volume of water = 30000 cm³/60 s = 500 cm³/s
Next, let's calculate the velocity of the water jet. We know that the cross-sectional area of the nozzle opening is 0.500 cm², so the volume of water flowing through the nozzle per second is:
Volume of water = velocity × cross-sectional area of the nozzle opening
We can rearrange this equation to solve for velocity:
velocity = Volume of water / cross-sectional area of the nozzle opening
velocity = 500 cm³/s / 0.500 cm² = 1000 cm/s
Now that we have the velocity of the water jet, we can calculate the horizontal distance it will travel in 1 second since the water is projected horizontally. The formula to calculate horizontal distance is:
distance = velocity × time
Since the time is 1 second, the horizontal distance is:
distance = 1000 cm/s × 1 s = 1000 cm
Therefore, the water will be projected over a horizontal distance of 1000 cm or 10 meters.