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Bess [88]
2 years ago
15

Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. If the nozzle exit diameter is 8 cm an

d the water flow rate is 12 m3/min, determine (a) the average water exit velocity and (b) the horizontal resistance force required of the firefighters to hold the nozzle.

Physics
2 answers:
raketka [301]2 years ago
8 0

Answer:

a) the average water exit velocity is 39.79m/s

b)the horizontal resistance force required of the firefighters to hold the nozzle is 7958N

Explanation:

Given that,

The exit diameter of the nozzle, D = 8cm = 0.08m

The volume water flow rate, Q = 12m³/min = 0.2m³/s

Density of water is 1000kg/m³

a)

V = \frac{Q}{A}

A = πd²/4

 = π(0.08)²/4

 = 5.03 × 10⁻³m₂

V = \frac{0.2}{5.03\times10^-^3} \\\\V = 39.79m/s

Therefore, the average water exit velocity is 39.79m/s

b)Calculate the mass flow rate using the equation below

m=pQ

where

p is the density of water

Q is the flow rate

m = 1000kg/m³ × 0.2m³/s

m = 200kg/s

Now the momentum equation for steady one dimensional flow

\sum \bar F=\sum_{out} \beta m \bar V - \sum_{in} \beta m \bar V

Now the momentum equation along x axis is

F_{Rx}=mV_{exit}\\\\=(200)\times(39.79)\\\\=7958N

Therefore, the horizontal resistance force required of the firefighters to hold the nozzle is 7958N

kobusy [5.1K]2 years ago
5 0

Answer:

Answer for the question is given in the attachment.

Explanation:

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Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
Natasha_Volkova [10]

The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

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where

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What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
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Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

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where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

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Answer:

uxjdkddoedlkfkkllllllllo

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