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motikmotik
3 years ago
9

You carry a fire hose up a ladder to a height of 10 m above ground level and aim the nozzle at a burning roof that is 9 m high.

You hold the hose horizontally and notice that the water strikes the roof at a horizontal distance of 7 m from where it exits the nozzle. The hose is connected to a large pressurized chamber in the fire truck 0.5 m above ground level. What is the pressure in the chamber
Physics
1 answer:
schepotkina [342]3 years ago
8 0

Answer:

The value is P_1  = 314645 \ Pa

Explanation:

From the question we are told that

The height is h_2  =  10  m

The height of the burning roof is k  =  9  m

The horizontal distance is d =  7 \  m

The height of the truck is h_1  =  0.5 \  m

Generally the time for the water to hit the roof from the hose is mathematically represented as

t =  \sqrt{\frac{2 *  (h_2 - k)}{g} }

=> t =  \sqrt{\frac{2 *  (10  - 9)}{9.8} }

=> t = 0.4518 \ s

Generally the velocity of the water is mathematically evaluated as

v_2  =  \frac{d}{t}

v_2  =  \frac{ 7}{0.4518}

v_2  =  15.5 \  m/s

Generally from Bernoulli's Equation we have that

P_1  + \frac{1}{2} v_1^2 * \rho + \rho *g *h_1  = P_2  + \frac{1}{2} v_2^2 * \rho + \rho *g *h_2

Here P_1 [\tex] is pressure in the chamber which we are to calculate , [tex]P_2 [\tex] is the atmospheric pressure with value  [tex]P_2 =  101325 \ Pa  [\tex] , [tex]v_1 [\tex] is the velocity of the water before it starts flowing with value [tex]v_1  =  0 m/s [\tex] , [tex]\rho [\tex]  is the density of water with value [tex]\rho =  1000 \ kg/m^3  [\tex] So         [tex]P_1  + \frac{1}{2} 0^2 * 1000 + 1000 *9.81 *0.5 = 101325  + \frac{1}{2}* 15.5^2* 1000 + 1000 *9.81 *10  

         P_1  = 314645 \ Pa        

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