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ser-zykov [4K]
3 years ago
15

The Kapandu boys dormitory in the UPNG is going up in flames. Rachel, a former student of Physics, who

Physics
1 answer:
Tom [10]3 years ago
4 0

Answer:

<em>1108.464 N of force</em>

Explanation:

diameter of water hose = 70 cm = 0.7 m

radius = 0.7/2 = 0.35 m

volumetric flow rate Q = 420 L/min

1 L = 0.001 m^3

1 min = 60 s

therefore,

Q = 420 L/min = (420 x 0.001)/60 = 0.007 m^3/s

Area A of fire hose = πr^{2} = 3.142 x 0.35^{2} = 0.38 m^2

<em>From continuity equation, Q = AV</em>

where V1 is the velocity of the water through the pipe, and A1 is the area of the pipe.

Q = A1V1

0.007 = 0.38V1

V1 = 0.007/0.38 = 0.018 m/s.

Nozzle diameter = 0.75 cm = 0.0075 m

radius = 0.00375

Area = πr^{2} = 3.142 x 0.00375^{2} = 4.42 x 10^{-5} m^2

velocity of water through the nozzle will be

V2 = Q/A2 = 0.007 ÷ (4.42 x 10^{-5}) = 158.37 m/s

From

<em>F = ρQ(v2 - v1)</em>

Where,

F = force exerted

p = density of water = 1000 kg/m^3

F = 1000 x 0.007 x (158.37 - 0.018) = <em>1108.464 N of force</em>

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(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

where, taking upward as positive direction:

v_y is the final vertical velocity

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v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

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v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

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As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

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