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Sedbober [7]
3 years ago
5

Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra

ys. Next, imagine an application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. Describe two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions.
Engineering
1 answer:
VashaNatasha [74]3 years ago
6 0

Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

1) struct array{

int *p;

}arr[1000];

2) struct array{

int *p;

}arr[1000];

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A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of
schepotkina [342]

Answer:

n = 2.36

Explanation:

The stress experimented by the circular bar is:

\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)

\sigma = 10.186\,kpsi

The safety factor is:

n = \frac{24\,kpsi}{10.186\,kpsi}

n = 2.36

5 0
3 years ago
Text that is located between <title >and </title > appears in the browser's_____​
Vinil7 [7]

Answer:

Tab title

Explanation:

The tab title can be defined as the title that shows up in a browser tab known as page title. The title only has texts, other tags within the texts are not considered.

8 0
2 years ago
Before you disconnect the service battery from the discharged battery, it is good practice to place a load across the
Lilit [14]

Answer:

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Explanation:

6 0
2 years ago
Read 2 more answers
Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,
Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take p_o = 880 kg/m³ and p_{glycerol = 1260 kg/m³    

 

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

Explanation:

Given that;

Inlet velocity of Glycerin, V_A = 6 m/s

Inlet velocity of oil, V_B = 3 m/s  

Density velocity of glycerin, p_{glycerol = 1260 kg/m³

Density velocity of glycerin, Take p_o = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, d_A = 100 mm = 0.1 m

Diameter of oil pipe, d_B = 80 mm = 0.08 m

Diameter of outlet pipe d_C = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

Q_A + Q_B = Q_C

A_Av_A + A_Bv_B = A_Cv_C

π/4 × (d_A)²v_A + π/4 × (d_B )²v_B = π/4 × (d_C)²v_C

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²v_C

0.04712 + 0.0150796 = 0.0113097v_C

0.0621996 = 0.0113097v_C

v_C = 0.0621996 / 0.0113097

v_C  = 5.5 m/s

Now we apply the mass flow rate condition

m_A + m_B = m_C

p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C  

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p  

72.6412 = 0.06220335p    

p = 72.6412 / 0.06220335

p =  1167.8 kg/m³  

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

4 0
2 years ago
When they say in the United States that a car’s tire is filled “to 32 lb,”​ they mean that <br>its internal pressure is 32 lbf/i
arsen [322]

Answer:

0.71 lbf

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.

P = 32 lbf/in² + 14.7 lbf/in²

P = 46.7 lbf/in²

Absolute temperature is in Kelvin or Rankine:

T = 75 + 459.67 R

T = 534.67 R

Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:

PV = nRT

(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)

n = 0.02442 lb-mol

The molar mass of air is 29 lbm/lb-mol, so the mass is:

m = (0.02442 lb-mol) (29 lbm/lb-mol)

m = 0.708 lbm

The weight of 1 lbm is lbf.

W = 0.708 lbf

Rounded to two significant figures, the weight of the air is 0.71 lbf.

3 0
2 years ago
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