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