A fire hose can project water at a rate of 6.00x10^2 litres per minute to a height of 40.0m. Find:

a) the velocity with which the water leaves the tube
b) the kinetic energy of the water as it leaves the hose
c) the kinetic energy of the water at 40.0m.

a) sqrt(2 g H),

where H = 40 m.

b) g*H per unit mass

c) zero. It temporarily stops moving at maximum height (if aimed straight up)

You don't need the liter/minute rate

To find the velocity with which the water leaves the tube, you can use the formula for projectile motion and the given data.

a) The formula for the velocity of an object in free fall motion is given by:

v = √(2gh)

where v is the velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

In this case, the height is 40.0m. Substitute these values into the formula:

v = √(2 × 9.8 m/s^2 × 40.0m)
= √(784 m^2/s^2)
≈ 28.0 m/s

The velocity with which the water leaves the tube is approximately 28.0 m/s.

Now, let's move on to part b.

b) The kinetic energy of an object can be calculated using the formula:

KE = 1/2 mv^2

where KE is the kinetic energy, m is the mass, and v is the velocity.

Since the mass of the water is not given, we can calculate it using the flow rate.

The given flow rate is 6.00 x 10^2 litres per minute. To convert this to cubic meters per second, divide by 1000 and then by 60:

6.00 x 10^2 L/min = (6.00 x 10^2 L/min) × (1 m^3 / 1000 L) × (1 min / 60 s)
= 10.0 m^3/s

As density (ρ) is mass (m) per unit volume (V), we need to know the density of water, which is approximately 1000 kg/m^3.

Using the formula:

m = ρV

where m is the mass, ρ is the density, and V is the volume, we can calculate the mass of the water:

m = (1000 kg/m^3) × (10.0 m^3/s)
= 10,000 kg/s

Substitute the values into the kinetic energy formula:

KE = 1/2 × (10,000 kg/s) × (28.0 m/s)^2
= 1/2 × 10,000 kg/s × 784 m^2/s^2
= 3,920,000 J

The kinetic energy of the water as it leaves the hose is approximately 3,920,000 Joules.

Now, let's move on to part c.

c) The kinetic energy of the water at a given height can be calculated using the conservation of energy.

The potential energy is given by the formula:

PE = mgh

where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.

Using the same mass value as before (10,000 kg/s) and the given height (40.0m), substitute these values into the formula:

PE = (10,000 kg/s) × (9.8 m/s^2) × (40.0 m)
= 3,920,000 J

The potential energy of the water at 40.0m is also approximately 3,920,000 Joules.

I hope this explanation helps! Let me know if you have any further questions.