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oksian1 [2.3K]
3 years ago
15

A water pipe having a 4.00 cm inside diameter carries water into the basement of a house at a speed of 1.00 m/s and a pressure o

f 167 kPa. The pipe tapers to 1.4 cm and rises to the second floor 7.8 m above the input point. What is the speed at the second floor
Physics
1 answer:
posledela3 years ago
8 0

Answer:

8.16\ \text{m/s}

Explanation:

d_1 = Initial diameter = 4 cm

v_1 = Initial velocity = 1 m/s

d_2 = Final diameter = 7.8 m

v_2 = Final velocity

A = Area = \pi\dfrac{d^2}{4}

From the continuity equation we get

A_1v_1=A_2v_2\\\Rightarrow \pi\dfrac{d_1^2}{4}v_1=\pi\dfrac{d_2^2}{4}v_2\\\Rightarrow v_2=\dfrac{d_1^2}{d_2^2}v_1\\\Rightarrow v_2=\dfrac{4^2}{1.4^2}\times 1\\\Rightarrow v_2=8.16\ \text{m/s}

The speed of water at the second floor is 8.16\ \text{m/s}.

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Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

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        D =150 10⁶ km (1000m / 1 km)

        D = 150 10⁹ m.

We must take the given angle to radians.

       1º  = 3600 arc s  

       π rad = 180º

       θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

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That angle is extremely small, so we can approximate the tangent to the angle

     

       θ = x / D

       x = θ D

       x = 4.85 10-6  150 109

       x = 727.5 103 m

       x = 727.5 km

4 0
3 years ago
You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie
Zinaida [17]

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance.  i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by =  $\frac{1}{0.06^2}$

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