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vredina [299]
4 years ago
7

Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have fou

r arithmetic operations which depend on the values of the selection variables as shown below. V1=0011, V0=0101 and output functions are as follows;
1- A+9's complement of B
2- A+B
3- A+10's complement of B
4- A+1 (add 1 to A)
(You can see question number 3 in the attached file)
Engineering
1 answer:
Bingel [31]4 years ago
4 0
Ucsaaaaauxx627384772938282’cc ed un e uff ridicolizzarla +golfista
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A hot plate with a temperature of 60 C, 50 triangular profile needle wings of length (54 mm), diameter 10 mm (k = 204W / mK) wil
frez [133]

Complete question is;

A hot plate with a temperature of 60 °C will be cooled by adding 50 triangular profile needle blades (k = 204 W/m.K) with a length of 54 mm and diameter 10 mm. According to the ambient temperature 20 °C and the heat transfer coefficient on the surface 20 W/m².K. Calculate,

a-) Wing efficiency

b-) Total heat transfer rate (W) from the wings,

c-) Calculate the effectiveness of a wing.

Answer:

A) Efficiency = 96.05 %

B) Total heat transfer rate = 166.68 W/m

C) Wing Effectiveness = 10.42

Explanation:

Please find attached explanation for all the answers given.

3 0
3 years ago
What is the next measurement after 2' -6" on the architect's scale?
Diano4ka-milaya [45]

Answer: I am not for sure

Explanation:

6 0
3 years ago
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa
Neko [114]

Answer:

Explanation:

The formula for critical stress is

\sigma_c=\frac{K}{Y\sqrt{\pi a} }

\sigma_c =\texttt{critical stress}

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

\sigma_c=\frac{54.8}{1\sqrt{\pi  \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa

The fracture will not occur because this material can handle a stress of 2186.20Mpa  before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm

4 0
3 years ago
the employer must comply with an employees or employee representative request to examine and copy abatement documents within
frez [133]
The employer must inform employees and their representatives of their right to examine and copy all abatement documents submitted to the Agency. ... The employer must comply with an employee's or employee representative's request to examine and copy abatement documents within 5 working days of receiving the request.
3 0
3 years ago
Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to su
Andrews [41]

Answer:

see explaination for all the answers and full working.

Explanation:

deflection=8P*DN/Gd^4

G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2

for outer spring,

deflection=8*3*50^3*5/(70*9^4)=32.66mm

for inner spring

deflection=8*3*30^3*10/(70*5^4)=148.11mm

max stress=k*8*P*C/(3.14*d^2)

for outer spring

c=50/9=5.55

k=(4c-1/4c-4)+.615/c=1.2768

max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2

for inner spring

c=6

k=1.2525

max stress=2.29KN/mm^2

6 0
4 years ago
Read 2 more answers
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