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3 years ago

It is required to design and implement: 1. A counter which counts from 0 to 255 with seven segment display.

Engineering

1 answer:

liubo4ka [24]3 years ago

5 0

Answer:

A counter which counts from 0 to 255 with seven segment display

Timer Mode Control (TMOD)

Explanation:

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Why do bridges collapse?

Reika [66]

Hi! bridges could have been collapse due to an error made by the engineers during construction.

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2 years ago

Read 2 more answers

Shortly after the introduction of a new coin, newspapers published articles claiming the coin is biased. The stories were based

garik1379 [7]

Answer:

(a) 0.12924

(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

Explanation:

(a)

n=200 for fair coin getting head, p= 0.5

Expectation = np =200*0.5=100

Variance = np(1 - p) = 100(1-0.5)=100*0.5=50

Standard deviation, $s = \sqrt {variance}=\sqrt {50}= 7.071068$

Z value for 108, $z =\frac {108-100}{7.071068}= 1.131371$

P( x ≥108) = P( z >1.13)= 0.12924

(b)

Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

3 0

3 years ago

A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De

andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes = {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes = {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes = {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0

2 years ago

Design a fundamental mode asynchronous finite state machine that accepts input pair, A and B. The AB input sequence 00, 01,11, 1

dybincka [34]

Answer:

See explaination

Explanation:

A Finite state machines can be synchronous or asynchronous. The operation of asynchronous state machines does not require a clock signal. An Asynchronous state machine is classified basically on their operating mode, such as the fundamental mode, pulse mode or burst mode. An asynchronous state machine can have stable and transient states.

Please kindly refer to attachment for a step by step solution.