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Ivahew [28]
3 years ago
12

A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide suffi

cient pressure in the pipes that deliver the water to customers. The drawing shows a spherical reservoir that contains 5.25  105 kg of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter of the delivery pipes

Physics
1 answer:
Mademuasel [1]3 years ago
4 0

The image of the water tower and the houses is in the attachment.

Answer: (a) P = 245kPa;

(b) P = 173.5 kPa

Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.

The pressure is calculated as: P = hρg

where

ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;

When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:

ρ = \frac{m}{V}

V = ρ/m

V = \frac{5.25.10^{5}}{10^{3}}

V = 525 m³

To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:

V = \frac{4}{3}.\pi.r^{3}

r = \sqrt[3]{ \frac{3}{4\pi } .V}

r = \sqrt[3]{\frac{525.3}{4\pi } }

r = 5.005 m

diameter = 2*r = 10.01m

(a) Height for House A:

h = 15 + 10.01

h = 25.01

P = hρg

P = 25.01.10³.9.8

P = 245.10³ Pa or 245kPa

(b) h = 25 - 7.3

h = 17.71

P = hρg

P = 17.71.1000.9.8

P = 173.5.10³ Pa or 173.5 kPa

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Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

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=>      \theta  = 613 \ revolutions

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Answer:

The correct option is A = 1960 N/m²

Explanation:

Given that,

Mass m= 20,000kg

Area A = 100m²

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