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2 years ago

What is the difference between a discrete and continuous system?

1 answer:

7 0

Answer:Discrete system are those system which change the state at the discrete point of time and continuous system are those which have state change at continuous period of time.

Explanation:

The major difference between discrete and continuous system is that, discrete system has state change over a discontinuous time period whereas the change of state in continuous system is over a continuous time period .
Variations can be found in the discrete system signal but in continuous system variation cannot be found input and output signal.
Example:-

Discrete system:-employees reporting at office at different time like

9:10am, 9:15am etc

<u> Continuous system</u>:-the water flow over a dam in particular quantity

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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

$= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}$

$= 1.751 \times 10^{-5} \% per hr$

4 0

2 years ago

In the engineering design and prototyping process, what is the advantage of drawings and symbols over written descriptions?

MrMuchimi

The advantages that can be associated to

drawings and symbols over written descriptions in engineering design and prototyping process are;

Communicate design ideas as well as technical information to engineers.

Symbols and drawings can be universal which means it is easy to interpret any where by professionals.

An engineering drawing serves as complex dimensional object and symbol use by engineer to communicate.

Drawings and symbols makes it easier to communicate design ideas and technical information to engineers and and how the process will go.

Therefore, drawings and symbols is universal to all engineer unlike written one.

Learn more at:

brainly.com/question/20925313?referrer=searchResults

4 0

2 years ago

A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the

padilas [110]

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

$A=\pi*D^2/4$

Replacing the diameter the area results:

$A= 17.76 in^2$

Therefore the the stress results:

$σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi$

5 0

2 years ago

Which of the following are true about the American Wire Gauge?

harina [27]

Answer:

A. smallest wire is No. 12

7 0

2 years ago

A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a

statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre = 183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0

2 years ago