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Ksju [112]
3 years ago
9

Given that the debouncing circuit is somewhat expensive in terms of hardware (2 NAND gates, 2 resistors, and a double-pole, sing

le throw switch), describe applications where you would require switch-debouncing circuits as well as applications where you would not need to include the additional hardware for switch debouncing (in other words, applications where you can tolerate switch bouncing). Note, you cannot use the clock and clear inputs of our lab as example applications; instead you need to think of other examples.
Engineering
1 answer:
kolezko [41]3 years ago
6 0

Explanation:

One of the common application of debouncing g circuit is in microprocessors or microcontrollers or FPGA's where fast processing is required. In such cases, it is extremely important that during the limited processing cycle, the signals remains valid without debouncinng. Because debouncing can complete impact the output of the controller.

A case where debouncing can be compromised where a system is run partially through human intervention or that has different indications for one operation.

For example in a car wash management system, where green and red lights are used to indicate if a car is being washed, green light will be on and then red light means that there no car in washing que

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Answer:

A) Cost

Explanation:

Chain drives have low maintenance cost.

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Shear strain can be expressed in units of either degrees or radians. a)True b)- False
ExtremeBDS [4]

Answer:

true

Explanation:

shear strain is define as the ratio of change in deformation to the original length perpendicular to the axes of member due to shear stress.

     ε    = deformation/original length

         

strain is a unit less quantity but shear stain is generally expressed in radians but it can also be expressed in degree.

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You are a planning aide on a team and are given the assignment of researching the historical significance and original blueprint
LenKa [72]
I’m guessing it will be the second one!
8 0
3 years ago
B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget
kap26 [50]

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = \frac{\pi }{4} * D^2 * l = \frac{\pi }{4} * 5.5^2 * 5 = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

4 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
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