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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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HOPE THIS HELPS!!!

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An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len
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Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

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3 years ago
the frequency of the middle b note on a piano is 493.88 hz. what is the wavelength of this note in centimeters? the speed of sou
vekshin1

Answer:

69.5

Explanation:

V=FX

343.06=493.88×X

X=343.06/493.88

X=0.695m

to cm is 69.5cm

5 0
2 years ago
A ______ is a closed loop containing a source of electrical energy and a load.
gavmur [86]

Answer:

d. circuit

Explanation:

the parts of the circuit consists of a load or resistance

7 0
3 years ago
A 295-kg object and a 595-kg object are separated by 4.10 m.
kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

F_{13}=\dfrac{Gm_1m_3}{r^2}

by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0
3 years ago
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