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3 years ago

Design a 3-bit binary counter using S-R flip flops.

Engineering

1 answer:

Helen [10]3 years ago

3 0

Answer:

This is an asynchrnous 3-bit counter. Just note that this design is different and works differently than its synchronous counterpart. It's an easier design than its synchronous counterpart, and is not as reliable because it has delays.

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

6 0

3 years ago

Given a two-dimensional steady inviscid air flow field with no body forces described by the velocity field given below. Assuming

kolbaska11 [484]

Answer:

the pressure gradient in the x direction = -15.48Pa/m

Explanation:

The concept of partial differentiation was used in the determination of the expression for u and v.
each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.

4 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questi

Schach [20]

Answer:

a) What is the surface temperature, in °C, after 400 s?

T (0,400 sec) = 800°C

b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s

c) What is the temperature, in °C, 1 mm from the surface after 400 s?

T (1 mm, 400 sec) = 798.35°C

Explanation:

oak initial Temperature = 25°C = 298 K

oak exposed to gas of temp = 800°C = 1073 K

h = 20 W/m².K

From the book, Oak properties are e=545kg/m³ k=0.19w/m.k Cp=2385J/kg.k

Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.

From energy balance: $\frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t$

Initial temperature wall = $T_i$

Surface temperature = T

Gas exposed temperature = $T_{\infty}$

6 0

3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

trasher [3.6K]

Answer:

The publication of a parody for commercial gain does not fall within the protection afforded by Section 107, as it is used for commercial gain.

Explanation:

<h2>PLEASE MARK AS BRAINLIEST!!!!!</h2>

4 0

2 years ago