How does a car batteray NOT die?

Simplify the following expressions, then implement them using digital logic gates. (a) f = A + AB + AC (b) f = AB + AC + BC (c)

Tema [17]

<u>Explanation</u>:

(a)

$f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A$

No gate is required to implement this function

(b)

$\begin{aligned}&\ f=A B+\bar{A} C+B C\\&\therefore f=A B+A C\end{aligned}$ $\begin{array}{l}(A B+\bar{A} C+B E=A B+\bar{A} C \\B C \text { is redendant })\end{array}$

Note: Refer the first image.

(c)

$\begin{aligned}f &=\overline{A+B}+A \bar{B}+B \bar{C} \\&=(\bar{A} \bar{B})+A \bar{B}+B \bar{C} \\&=\bar{B}[A+\bar{A}]+B \bar{C} \\& F=\bar{B}+B \bar{C} =\bar{B}+\bar{C}\end{aligned}$

Note: Refer the second image

(d)

$\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}$

Note: Refer the third image

(e)

$\begin{aligned}f=& A \bar{B}+\bar{B} C+A \bar{B} \\&=\bar{B}[A+\bar{A}+c] \\&=\bar{B}[1+C]\end{aligned}$

$f=\frac{}{B}$

(f)

$\begin{aligned}f &=A B C+A B D+A B C \\&=A B[C+C]+A B D \\&=A B+A B D \\&=B[A+A D] \\&=B[A+D] \\\therefore & A=B[A+D]\end{aligned}$

Note: Refer the fourth image

6 0

3 years ago

In an apartment the interior air temperature is 20°C and exterior air temperatures is 5°C. The wall has inner and outer surface

ELEN [110]

Answer:

20 W/m², 20 W/m², -20 W/m²

Yes, the wall is under steady-state conditions.

Explanation:

Air temperature in room = 20°C

Air temperature outside = 5°C

Wall inner temperature = 16°C

Wall outer temperature = 6°C

Inner heat transfer coefficient = 5 W/m²K

Outer heat transfer coefficient = 20 W/m²K

Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)

q = hΔT

Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)

⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²

Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)

⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²

Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)

⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²

Here the magnitude of the heat flux are same so the wall is under steady-state conditions.

7 0

3 years ago

For a metal that has a yield strength of 690 MPa and a plane strain fracture toughness (KIc) of 32 MPa-m1/2, compute the minimum

Digiron [165]

Answer:

the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

Explanation:

Given the data in the question;

yield strength σ $_y$ = 690 Mpa

plane strain fracture toughness K $_{Ic$ = 32 MPa- $m^{1/2$

minimum component thickness for which the condition of plane strain is valid = ?

Now, for plane strain conditions, the minimum thickness required is expressed as;

t ≥ 2.5( K $_{Ic$ / σ $_y$ )²

so we substitute our values into the formula

t ≥ 2.5( 32 / 690 )²

t ≥ 2.5( 0.0463768 )²

t ≥ 2.5 × 0.0021508

t ≥ 0.005377 m or 5.38 mm

Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

7 0

3 years ago

She spent her time reading stories about people who fought in the Revolution.

Sonbull [250]

She spent her time reading stories about people who fought in the Revolution.

4 0

3 years ago

Which of the following have had a significant impact on the automotive industry?

WARRIOR [948]

All of the above (hope that helps!)

6 0

3 years ago