Provided the diameter of the fire hose is 7.0 cm, and the water running through the hose flows at 420 L/min, how much force does she need to hold the nozzle, if its diameter is 0.75 cm?

Respuesta :

Answer:

the amount of force  needed to hold the nozzle is 1100 N

Explanation:

Given that;

The diameter of the fire horse is 7.0cm

The radius of the fire horse is d/2 = 7.0/2 = 3.5 cm = 0.035 m

the water running through the hose flows at 420 L/min = 0.42 m³/ 60 sec

The diameter of the nozzle = 0.75 cm

The radius of the nozzle = d/2 = 0.75/2 = 0.375 cm = 0.00375 m

Amount of force needed to hold the nozzle = ??

Using equation of continuity product

[tex]A_1v_1 = A_2v_2[/tex]

where

[tex]A_1[/tex] = cross sectional area of the hose

[tex]v_1[/tex] = velocity of water flow at hose

[tex]A_2[/tex] = cross sectional area of the nozzle

[tex]v_2[/tex] = velocity of water flow at nozzle

Making [tex]v_1[/tex] the subject of the formula; we have

[tex]v_1 = \dfrac{A_2v_2}{A_1}[/tex]

Also making [tex]v_2[/tex] the subject of the formula:

[tex]v_2 = \dfrac{A_1v_1}{A_2}[/tex]

Also; from newton's law; we all know that F = ma

where ;

m = mass

a = acceleration

[tex]a = \dfrac{\Delta v}{t} \\ \\ a = \dfrac{v_2-v_1}{t}[/tex]

So ;

[tex]F = m \dfrac{v_2-v_1}{t}[/tex]

Also ; mass = density × volume

[tex]m =\rho *V[/tex]

[tex]F = \rho V \dfrac{v_2-v_1}{t}[/tex]

[tex]F = \dfrac{\rho V}{t}[ \dfrac{A_1v_1}{A_2}- \dfrac{A_2v_2}{A_1}][/tex]

Replacing [tex]A_2v_2[/tex]  by [tex]A_1v_1[/tex] from above ; so

[tex]F = \dfrac{\rho V}{t}[ \dfrac{A_1v_1}{A_2}- \dfrac{A_1v_1}{A_1}][/tex]

[tex]F = \dfrac{\rho V}{t} A_1 v_1 [ \dfrac{1}{A_2}- \dfrac{1}{A_1}][/tex]

where the product of cross section area of velocity is equal to the volume of water and its time flow;

SO;

[tex]A_1v_1 = \dfrac{V}{t}[/tex]

Replacing that into what we have above; we have:

[tex]F = \dfrac{\rho V}{t} \dfrac{V}{t} [ \dfrac{1}{A_2}- \dfrac{1}{A_1}][/tex]

[tex]F ={\rho }( \dfrac{V}{t})^2 [ \dfrac{1}{A_2}- \dfrac{1}{A_1}][/tex]

Where;

A = πr²

[tex]F ={\rho }( \dfrac{V}{t})^2 [ \dfrac{1}{\pi r^2_2}- \dfrac{1}{\pi r^2_1}][/tex]

[tex]F ={(1000*10^3 \ kg/m^3) }( \dfrac{0.42 m^3 }{60 sec})^2 [ \dfrac{1}{\pi (0.00375)^2}- \dfrac{1}{\pi (0.035)^2}][/tex]

F = 1100 N

Thus; the amount of force  needed to hold the nozzle is 1100 N