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konstantin123 [22]
3 years ago
5

For the data points (1,1),(2,1/2),(3,1/3),(4,1/4), finde the natural cubic spline.

Engineering
1 answer:
kow [346]3 years ago
7 0

Answer:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

Explanation:

A cubic has the form:

y = ax³ + bx² + cx + d

Given four points, we can write a system of equations:

1 = a + b + c + d

1/2 = 8a + 4b + 2c + d

1/3 = 27a + 9b + 3c + d

1/4 = 64a + 16b + 4c + d

Solving this algebraically would be time-consuming, but we can use matrices to make it easy.

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right]\left[\begin{array}{cccc}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right]

First, we find the inverse of the coefficient matrix.  This is messy to do by hand, so let's use a calculator:

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right] ^{-1} =-\frac{1}{12}\left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]

Now we multiply by the solution matrix (again using a calculator):

-\frac{1}{12} \left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right] =\left[\begin{array}{cccc}-1/24\\5/12\\-35/24\\25/12\end{array}\right]

So the cubic is:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

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lidiya [134]

Answer:

Q = 62    ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol

T_1 = 27^0 \ C = ( 27+273)K =  300 K

P_1 = 200 \ kPa

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if V_1 = x\\V_2 = 2V_1

Using Charles Law; since pressure is constant

V \alpha T

\frac{V_2}{V_1}  =\frac{T_2}{T_1}

\frac{2V_1}{V_1}  =\frac{T_2}{300}

T_2 = 300*2\\T_2 = 600

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg  × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg  × 40

mass of He = 280 kg

Now; the amount of  Heat  Q transferred = m_{He}Cp_{He} \delta T  + m_{Ar}Cp_{Ar} \delta T

From gas table

Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar}  = 0.5203 \  kJ/Kg/K

∴ Q = 12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)

Q = 62.389 *10^6

Q = 62 MJ

Q = 62    ( since we are instructed not to include the units in the answer)

5 0
3 years ago
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"Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."
kirill [66]

Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

Java <em>println() </em>method can be used to display any string on the console terminal. We can use <em>println()</em> method to output the value held by variable <em>numItems.</em> The <em>numItems </em>is passed as the input parameter to <em>println()</em> and this will output the value of <em>numItems</em> to console terminal and at the same time the output with be ended with a newline automatically.  

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For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time
max2010maxim [7]

Answer with Explanation:

The general equation of simple harmonic motion is

x(t)=Asin(\omega t+\phi)

where,

A is the amplitude of motion

\omega is the angular frequency of the motion

\phi is known as initial phase

part 1)

Now by definition of velocity we have

v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )

part 2)

Now by definition of acceleration we have

a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )

part 3)

The angular frequency is related to Time period 'T' asT =\frac{2\pi }{\omega }

where

\omega is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is A\omega ^{2} and A\omega respectively.

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What is the built-in pollution control system in an incinerator called
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Explanation:

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An asphalt concrete mixture includes 94% aggregate by weight. The specific gravities of aggregate and asphalt are 2.65 and 1.0,
garik1379 [7]

Answer:

2.0%

Explanation:

Percentage of aggregate = 94%

Specific gravity = 2.65

Specific gravity of asphalt = 1.9

Density of mix = 147pcf = 147lb/ft³

Total weight of mix: (volume = 1ft³)

= (147lb/ft³)(1ft³)

= 147lb

Percentage weight of asphalt in<u> mix:</u>

100% - 94%

= 6%

Weight of asphalt binders

= 6% x 147lb

= 8.82lb

Weight of aggregate in mix:

= 94% x 147

= 138.18lb

Specific weight of asphalt binder:

(Gab)(Yw)

Yw = specific Weight of water

= 62.4lb

Gab = specific gravity of asphalt binder

= 1.0

(62.4lb)(1.0)

= 62.4 lb/ft³

Volume of asphalt in binder:

8.82/62.4

= 0.14ft³

Specific weight of binder in mix:

2.65 x 62.4lb/ft³

= 165.36 lb/ft³

Volume of aggregate:

= 138.18/165.36

= 0.84ft³

Volume of void in the mix:

1ft³ - 0.84ft³ - 0.14ft³

= 0.02ft³

<u>The percentage of void in total mix:</u>

VTM = (0.02ft³/1ft³)100

= 2.0%

8 0
3 years ago
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