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katovenus [111]
3 years ago
11

Make two lists of applications of matrices, one for those that require jagged matrices and one for those that require rectangula

r matrices. Now argue whether just jagged, just rectangular, or both should be included in a programming language.
Engineering
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

Explanation:

You can utilize barbed clusters to store inadequate grids. On the off chance that there are a great many lines yet each line has just 4 or 5 associations with different segments, at that point as opposed to utilizing a 1000x1000 cluster you can utilize a 1000 line rough exhibit while you simply store the components that the present section has association with another segment. Other utilization can be done on account of query tables. Query tables will be tables which have different qualities concerning a solitary key where the quantity of qualities isn't fixed. Aside from this, barbed clusters have an exceptionally set number of utilization cases. Multidimensional exhibits then again have plenty of utilizations. It is utilized to store a great deal of information reliably on the grounds that the greater part of the information is put away is steady concerning which section compares to what information. Aside from that it very well may be utilized to make thick diagrams or sparse(not effective), plotting information. Another utilization case would be used as an impermanent stockpiling for the figurings that need to tail them and utilize the past information like in powerful programming.

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A dual-fluid heat exchanger has 10 lbm/s water entering at 100 F, 20 psia and leaving at 50 F, 20 psia. The other fluid is glyco
kakasveta [241]

Answer:

Rate of internal heat transfer = 23.2 Btu/Ibm

mass flow rate = 21.55 Ibm/s

Explanation:

using given data to obtain values from table F7.1

Enthalpy of water at temperature of 100 F = 68.04Btu/Ibm

Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm

from table F.3

specific constant of glycerin C_{p} = 0.58 Btu/Ibm-R

<u>The rate of internal heat transfer ( change in enthalpy ) </u>

h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )

where ; T4 = 50 F

             T3 = 10 F

             Cp = 0.58 Btu/Ibm-R

substitute given values into equation 1

change in enthalpy ( h4 - h3 ) = 23.2 Btu/Ibm

<u>Determine mass flow rate of glycol</u>

attached below is the detailed solution

mass flow rate of glycol = 21.55 Ibm/s

3 0
3 years ago
the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when
pshichka [43]

Answer:

80.16 m/s^2

at t=2 s

x=42.3 m

y=16 m

z=14 m

Explanation:

solution

The x,y,z components of the velocity are donated by the i,j,k vectors.

v_{x}=16t^{2}  \\v_{y}=4t^{3}\\v_{z}=5t+2

acceleration is a derivative of velocity with respect to time.

a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5

evaluate acceleration at 2 seconds

a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2

position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.

x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\

3 0
2 years ago
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
Identify factors that can cause a process to become out of control. Give several examples of such factors.
Oliga [24]

Answer:

Explained

Explanation:

This situation can occur because of various factors such as:

  • Gradual deterioration of lubrication and coolant.
  • change of environmental condition such as temperature, humidity, moisture, etc.
  • Change in the properties of incoming raw material
  • An increase or decrease in the temperature of the heat treating operation
  • Debris interfering with the manufacturing process.
4 0
2 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
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