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4 years ago

Write an ALP to separate odd and even numbers from an array of N numbers; arrange odd

Engineering

1 answer:

Marta_Voda [28]4 years ago

6 0

Below is the program to separate odd and even numbers

Explanation:

L1:

mov ah,00

mov al,[BX]

mov dl,al

div dh

cmp ah,00

je EVEN1

mov [DI],dl

add OddAdd,dl

INC DI

INC BX

Loop L1

jmp CAL

EVEN1:

mov [SI],dl

add Even Add,dl

INC SI

INC BX

Loop L1

CAL:

mov ax,0000

mov bx,0000

mov al,OddAdd

mov bl,EvenAdd

MOV ax,4C00h

int 21h

end

The above program separates odd and even numbers from the array using 8086 microprocessor. It has odd numbers in 2000h and even numbers in 3000h.

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3 years ago

What are the characteristics of Polyamide?

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3 years ago

A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$

r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Now by putting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

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4 years ago

Which of the following is true of the difference between film acting and stage acting?

vladimir1956 [14]

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Further Explanation:

There are numerous other ways that film and stage acting is different. In addition to the audiences, there is the extreme action shots that a film actor can accomplish where a stage actor cannot.

Other notable differences are;

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Learn more about film actors at brainly.com/question/3511000

#LearnwithBrainly

4 0

3 years ago

A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and

ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

$h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg$

The quality at state 4 is determined from the condition $s_{4} =s_{3}$ and the entropies of the components at the condenser pressure taken from table:

$q_{4} =\frac{s_{4}-s_{liq50} }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935$

The enthalpy at state 4 then is:

$h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\$

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:

$h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg$

The enthalpy at state 2 is determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

=346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349$

Part B

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:

$h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg$

The enthalpy at state 2 is then determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

=299.79 kJ/kg

The thermal efficiency in this case then is:

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345$

8 0

3 years ago