A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on about 61.0% of the roof. A typical annual rainfall is about 14.0 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water in a full tank of that size weigh (in N and in lbf)?

Answers

Answer 1
Answer:

Answer:

volume  = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

volume  =  A'* H

where A' = 61% of A

              = 0.61* 2667 = 1626.87 sq ft

volume  =  1626.87 * ((14)/(12) ft)

               =1898.015 ft^3

in\ m^3 = ( 1898.015)/(35.315) =   53.7457 m^3

in\ gallon = 1898.015 * 7.481 = 14198.138 gallon

weight = \rho Vg

       = 1000* 53.74* 9.8

             =526652 N

In\ lbf =  (526652)/(4.448) = 118396.08 lbf


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An automobile engine consumes fuel at a rate of 27.4 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.73 g/cm3, deter- mine the efficiency of this engine in percentage(

Answers

Answer:

The efficiency of the engine is 22.5%.

Explanation:

Efficiency = power output ÷ power input

power output = 55 kW

power input = specific energy×volumetric flow rate×density

specific energy = 44,000 kJ/kg

volumetric flow rate = 27.4 L/h = 27.4 L/h × 1000 cm^3/1 L × 1 h/3600 s = 7.61 cm^3/s

density = 0.73 g/cm^3 = 0.73 g/cm^3 × 1 kg/1000g = 7.3×10^-4 kg/cm^3

power input = 44,000 kJ/kg × 7.61 cm^3/s × 7.3×10^-4 kg/cm^3 = 244.4332 kJ/s = 244.4332 kW

Efficiency = 55 ÷ 244.4332 = 0.225 × 100 = 22.5%

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.

Answers

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively.

Answers

Answer:

L = 75.25 mm

Explanation:

First we need to find the lateral strain:

Lateral Strain = Change in Diameter/Original Diameter

Lateral Strain = (20.025 mm - 20 mm)/20 mm

Lateral Strain = 1.25 x 10⁻³

Now, we will find the Poisson's Ratio:

Poisson's Ratio = (E/2G) - 1

where,

E = Elastic Modulus = 105 GPa

G = Shear Modulus = 39.7 GPa

Therefore,

Poisson's Ratio = [(105 GPa)/(2)(39.7 GPa)] - 1

Poisson's Ratio = 0.322

Now, we find longitudinal strain by following formula:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

Longitudinal Strain = - Lateral Strain/Poisson's Ratio

Longitudinal Strain = - (1.25 x 10⁻³)/0.322

Longitudinal Strain = - 3.87 x 10⁻³

Now, we can fin the original length:

Longitudinal Strain = Change in Length/L

where,

L = Original Length = ?

Therefore,

- 3.87 x 10⁻³ = (74.96 mm - L)/L

(- 3.87 x 10⁻³)(L) + L = 74.96 mm

0.99612 L = 74.96 mm

L = 74.96 mm/0.99612

L = 75.25 mm

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating

Answers

Answer:

The answer is below

Explanation:

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Given that:

number of poles (p) = 4, frequency (f) = 60 Hz

1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:

n_s=(120f)/(p)=(120*60)/(4)=1800\ rpm

2) The slip (s) = 0.05

The speed of the motor (n) is the speed of the rotor, it is given as:

n=n_s-sn_s\n\nn=1800-0.05(1800)\n\nn=1800-90\n\nn=1710\ rpm

3) s = 0.04

The rotor frequency is the product of the supply frequency and slip it is given as:

f_r=sf\n\nf_r=0.04*60\n\nf_r=2.4\ Hz

4) At standstill, the motor speed is zero hence the slip = 1:

s=(n_s-n)/(n_s)\n \nn=0\n\ns=(n_s-0)/(n_s)\n\ns=1

The rotor frequency is the product of the supply frequency and slip it is given as:

f_r=sf\n\nf_r=1*60\n\nf_r=60\ Hz

At full load, a commercially available 100hp, three phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is supplied from a three-phase outlet with a line voltage rating of 208V.a. What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b. Calculate the reactive power supplied to the motor.

Answers

Answer:

I = Line Current = 242.58 A

Q = Reactive Power = 41.5 kVAr

Explanation:

Firstly, converting 100 hp to kW.

Since, 1 hp = 0.746 kW,

100 hp = 0.746 kW x 100

100 hp = 74.6 kW

The power of a three phase induction motor can be given as:

P_(in)  = √(3) VI Cos\alpha\n

where,

P in = Input Power required by the motor

V = Line Voltage

I = Line Current

Cosα = Power Factor

Now, calculating Pin:

efficiency = \frac{{P_(out)} }{P_(in) }\n0.97 = (74.6)/(P_(in) ) \nP_(in) = (74.6)/(0.97)\n  P_(in) = 76.9 kW

a) Calculating the line current:

P_(in) = √(3)VICos\alpha   \n76.9 * 1000= √(3)*208*I*0.88\nI = (76.9*1000)/(√(3)*208*0.88 )\nI =   242.58 A

b) Calculating Reactive Power:

The reactive power can be calculated as:

Q = P tanα

where,

Q = Reactive power

P = Active Power

α = power factor angle

Since,

Cos\alpha =0.88\n\alpha =Cos^(-1)(0.88)\n\alpha=28.36

Therefore,

Q = 76.9 * tan (28.36)\nQ = 76.9 * (0.5397)\nQ = 41. 5 kVAr

Karen just checked her bank balance and the balance was 1111.11. She is kind of freaking out and saved the receipt, thinking that it must mean something. Karen is probably just suffering from

Answers

Answer:

chronic stoner syndrome

Explanation:

"the universe just sends us messages sometimes mannnn, you just have to be ready to listen to them" lol

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