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alexandr1967 [171]
3 years ago
7

A pipe leads from a storage tank on the roof of a building to the ground floor. The absolute pressure of the water in the storag

e tank where it connects to the pipe is 3.0×105Pa, the pipe has a radius of 1.0 cm where it connects to the storage tank, and the speed of flow in this pipe is 1.6 m/s. The pipe on the ground floor has a radius of 0.50 cm and is 9.0 m below the storage tank. Find the speed of flow in the pipe on the ground floor.
Engineering
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

6.4 m/s

Explanation:

From the equation of continuity

A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively

A1=\pi (r1)^{2}

A2=\pi (r2)^{2}

where r1 and r2 are radius of inlet and outlet respectively

v1 is given as 1.6 m/s

Therefore

\pi (0.01)^{2}\times 1.6 = \pi (0.005)^{2}v2

V2=\frac {\pi (0.01)^{2}\times 1.6}{\pi (0.005)^{2}}=6.4 m/s

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An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs
Greeley [361]

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h=110\times \frac{5}{18}=30.56 m/s

Where 1 km/h=5/18 m/s

Height , h_2=40 m

g=9.81 m/s^2

Final speed, v=0

Change in kinetic energy,\Delta K.E=\frac{1}{2}m(v^2-u^2)

\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J

\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ

Where 1 KJ=1000 J

Change in potential energy,\Delta P.E=mgh(h_2-h_1)

Initially height, h1=0

Using the formula

\Delta P.E=1100\times 9.81(40-0)

\Delta P.E=431640J

\Delta P.E=431.64KJ

6 0
3 years ago
Summarize key
BlackZzzverrR [31]

Answer:

what are is ethiopia cultural ?

7 0
2 years ago
Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
Nutka1998 [239]

Answer:

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given :  x + 2x + 2x = 0   for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = \sqrt{\frac{k}{m} }  = \sqrt{2}  

next we determine the damping condition using the damping formula

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

6 0
2 years ago
What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
yanalaym [24]

Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

7 0
3 years ago
/* Function findBestVacation * duration: number of vacation days * prefs: prefs[k] indicates the rate specified for game k * pla
alexira [117]

Answer:

This is the C++ code for the above problem:

#include<bits/stdc++.h>

using namespace std;

int computeFunLevel(int start, int duration, int prefs[], int ngames, int plan[]) {

if(start + duration > 365) { //this is to check wether duration is more than total no. of vaccation days

return -1;

}

int funLevel = 0;

for(int i=start; i<start+duration; i++) { //this loop runs from starting point till

//start + duration to sum all the funlevel in plan.

funLevel = funLevel + prefs[plan[i]];

}

return funLevel;

}

int findBestVacation(int duration, int prefs[], int ngames, int plan[]) {

int max = 0, index = 0, sum = 0 ;

for(int i=1; i<11; i++){ //this loop is to run through whole plan arry

sum = 0; //sum is initialized with zero for every call in plan ,

//in this case loop should run to 366,but for example it is 11

//as my size of plan array is 11

for(int j=0; j<duration; j++) { // this loop is for that index to index+duration to calc

//fun from that index

sum = sum + prefs[plan[i]];

}

if(sum>max) { //this is to check max funlevel and update the index from which max fun can be achieved

max = sum;

index = i;

}

}

return index;

}

int main() {

int ngames = 5;

int prefs[] = { 1,2,0,5,2 };

int plan[] = { 0,2,0,3,3,4,0,1,2,3,3 };

int start = 1;

int duration = 4;

cout << computeFunLevel(start, duration, prefs, ngames, plan) << endl;

cout << computeFunLevel(start, 555, prefs, ngames, plan) << endl;

cout << findBestVacation(4, prefs, ngames, plan) << endl;

}

The screen of the program is given below.

3 0
3 years ago
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