For a super soaker water gun, a pump handle is moved back and forth to build up pressure in the water reservoir. The water is released by pulling a trigger and shooting the water a significant distance. Assuming that you can create an absolute pressure of 8 atm in the reservoir: a) What is the velocity at which the water leaves the gun?
b) If the water exits the gun through a hole with a radius of 1-mm, what is the volume rate of flow in m3/s?
c) If the water gun is fired horizontally and held 1.2 meters above the ground, where does the water hit the ground? Pressure 8 cm water

a) The velocity at which the water leaves the gun = 37.66 m/s

b) The volume rate of flow = (1.183 × 10⁻⁴) m³/s

c) The water hits the ground 18.64 m from the point where the water gun was shot.

Step-by-step explanation:

a) Using Bernoulli's equation, an equation that is based on the conservation of energy.

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

The two levels we are considering is just inside the water reservoir and just outside it.

ρgh is an extension of potential energy and since the two levels are at the same height,

ρgh₁ = ρgh₂

Bernoulli's equation becomes

P₁ + ½ρv₁² = P₂ + ½ρv₂²

P₁ = Pressure inside the water reservoir = 8 atm = 8 × 101325 = 810,600 Pa

ρ = density of water = 1000 kg/m³

v₁ = velocity iof f water in the reservoir = 0 m/s

P₂ = Pressure outside the water reservoir = atmospheric pressure = 1 atm = 1 × 101325 = 101,325 Pa

v₂ = velocity outside the reservoir = ?

810,600 + 0 = 101,325 + 0.5×1000×v₂²

500v₂² = 810,600 - 101,325 = 709,275

v₂² = (709,275/500) = 1,418.55

v₂ = √(1418.55) = 37.66 m/s

b) Volumetric flowrate is given as

Q = Av

A = Cross sectional Area of the channel of flow = πr² = π×(0.001)² = 0.0000031416 m²

v = velocity = 37.66 m/s

Q = 0.0000031416 × 37.66 = 0.0001183123 m³/s = (1.183 × 10⁻⁴) m³/s

c) If the height of gun above the ground is 1.2 m. Where does the water hit the ground?

The range of trajectory motion is given as

R = vT

v = horizontal component of the velocity = 37.66 m/s

T = time of flight = ?

But time of flight is given as

T = √(2H/g) (Since the initial vertical component of the velocity = 0 m/s

H = 1.2 m

g = acceleration due to gravity = 9.8 m/s²

T = √(2×1.2/9.8) = 0.495 s

Range = vT = 37.66 × 0.495 = 18.64 m

Hope this Helps!!!

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.