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slavikrds [6]
3 years ago
9

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the

water speed doubles and the cross-sectional area of the pipe triples, what happens to the volume flow rate of the water passing through it?
Engineering
1 answer:
Arlecino [84]3 years ago
8 0

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

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garri49 [273]

Answer:

c. and d

Explanation:

As a whistle-blower, one of your aim is to guide against unethical dealings of other people , hence you are creating an environment that uphold ethical conduct,

In addition, whistle-blowing will disclose all imminent dangers to the software community thereby preventing security breaches.

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3 years ago
Write a program that uses a function called Output_Array_Info. Output_Array_Info Properties: Input Parameters: 1. A pointer to a
Artyom0805 [142]

Answer:

C++ code explained below

Explanation:

Please note the below program has been tested on ubuntu 16.04 system and compiled using g++ compiler. This code will also work on other IDE's

-----------------------------------------------------------------------------------------------------------------------------------

Program:

-----------------------------------------------------------------------------------------------------------------------------------

//header files

#include<iostream>

//namespace

using namespace std;

//function defintion

void Output_Array_Info(int *array_ptr, int size)

{

//display all array elements

cout<<"Array elements are: "<<endl;

for(int i =0; i<size; i++)

{

cout<<*(array_ptr+i)<<endl;

}

//display address of each element

cout<<endl<<"memory address of each array elemnt is: "<<endl;

for(int i =0; i<size; i++)

{

cout<<array_ptr+i<<endl;

}

}

//start of main function

int main()

{

//pointer variables

int *pointer;

//an array

int numbers[] = { 5, 7, 9, 10, 12};

//pointer pointing to array

pointer = numbers;

//calculate the size of the array

int size = sizeof(numbers)/sizeof(int);

//call to function

Output_Array_Info(numbers, size);

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8 0
3 years ago
A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

dQ = m \times c_p \times (T_2 -T_1)

For ideal gas, c_p = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

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The answer is 32.0 grams
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julsineya [31]

Answer:

Walk to saftey!

Explanation:

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