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3 years ago

14

Suppose we want to determine how many of the bits in a twelve-bit unsigned number are equal to zero. Implement the simplest circ

uit that can accomplish this task.

1 answer:

timurjin [86]3 years ago

8 0

Answer:

See the picture for circuit figure.

Explanation:

First load 12 bit unsigned number on shift register and apply clock tick to increment counter from LSB bit.

Logic 1 is shifted into register to ensure counter doesn't count more than 12 bits.

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Disconnecting a circuit while in operation can create a voltage blank

zlopas [31]

Answer:

what is the question

Explanation:

confused

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3 years ago

A simply supported wood roof beam is loaded with single point dead and roof live loads applied at midspan (PD = 400 lb, PLr = 16

Lynna [10]

Answer:mold i belive

Explanation:

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3 years ago

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

$= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

$H=mh_x$

$= 0.5 \times 2045.83$

= 1022.91 kJ

7 0

3 years ago

Solve using Matlab the problems:

Firlakuza [10]

Answer:

Explanation:

% Clears variables and screen

clear; clc

% Asks user for input

n = input('Total number of objects: ');

r = input('Size of subgroup: ');

% Computes and displays permutation according to basic formulas

p = 1;

for i = n - r + 1 : n

p = p*i;

end

str1 = [num2str(p) ' permutations'];

disp(str1)

% Computes and displays combinations according to basic formulas

str2 = [num2str(p/factorial(r)) ' combinations'];

disp(str2)

=================================================================================

Example: check

How many permutations and combinations can be made of the 15 alphabets, taking four at a time?

The answer is:

32760 permutations

1365 combinations

==================================================================================

7 0

3 years ago

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0

3 years ago