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I rent a small high pressure water sprayer to clean the outside of my house. The sprayer works like a super soaker with a hose between the tank and the nozzle. You pump up the pressure in the tank with a handle and then open the nozzle to squirt water, or cleaning fluid, or insecticide etc... The small water sprayer tank sits on the ground (since the tank is small you can assume that all the water in the tank is at the same height....the height of the ground). The top of my house walls are 8m above the ground. I hold the nozzle of the hose 1.6m above the ground. (Ignore friction in Questions 1-3)What is the minimum additional pressure (above atmospheric pressure) necessary inside the sprayer if I wish to clean all the way to the top of the walls?(Hint: what physical principle needed to answer this question and what equation?)8 Pa800 Pa80000 Pa8000000 Pa2.Assume that the sprayer has the internal pressure calculated in Question 1. I’m still holding the nozzle 1.6m above the ground. What is the velocity of water leaving the nozzle of the sprayer?1.13 m/s11.3 m/s1.2 m/s12 m/s3.My child holds the nozzle at ground level (instead of 1.6m above the ground) what is the velocity of the water leaving the sprayer? (the sprayer still has the same pressure inside the tank calculated in Question 1)9.9 m/s0 m/s^212.5 m/s15.4 m/s13.6 m/s

Answer :

opudodennis

Answer:

1. 80,000 Pa

2. 11.3 m/s

3. 12.5 m/s

Explanation:

Question 1

Pressure, [tex]P=hg\rho[/tex]

Where h is the height that water is to reach, g is gravitational constant and [tex]\rho[/tex] is the density, in this case, we assume [tex]\rho[/tex] of pure water as [tex]1000 Kg/m^3[/tex]

Assuming [tex]g=10 m/s^{2}[/tex]

P=8*10*1000=80000 Pa

Question 2

Pressure can also be found by the formula

[tex]P=0.5v^{2}\rho[/tex] where v is the velocity

Equating the new formula of pressure to the formula used in question 1 above

[tex]P=0.5v^{2}\rho=hg\rho[/tex]

Notice that [tex]\rho[/tex] is common hence

[tex]0.5v^{2}=hg[/tex]

Making V the subject of the formula

[tex]v^{2}=2hg[/tex]

[tex]v=\sqrt 2hg[/tex]

In this case, h=8-1.6=6.4m and taking g as 10 m/s^{2}

[tex]v=\sqrt 2*10*6.4=11.3137085  m/s[/tex]

Rounding off to 1 decimal place

v=11.3 m/s

Question 3

As already illustrated

[tex]v=\sqrt 2hg[/tex]

Taking g as 9.8 and h now is 8m

[tex]v=\sqrt 2*8*9.8[/tex]

v=12.52198067

Rounding off to 1 decimal place

v=12.5 m/s

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