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fredd [130]
3 years ago
9

Why did the dodge demon cross the road

Engineering
2 answers:
fgiga [73]3 years ago
5 0

Answer:

to get to the other side

Zanzabum3 years ago
4 0
To get to the other side ‼️
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Design a 10-to-4 encoder with inputs in the l-out-of-10 code and outputs in a code like normal BCD except that input lines 8 and
zvonat [6]

Answer:

See image attached.

Explanation:

This device features priority encoding of the inputs

to ensure that only the highest order data line is en-

coded. Nine input lines are encoded to a four line

BCD output. The implied decimal zero condition re-

quires no input condition as zero is encoded when

all nine datalinesare athigh logic level. Alldata input

and outputs are active at the low logic level. All in-

puts are equipped with protection circuits against

static discharge and transient excess voltage.

6 0
4 years ago
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a
Fed [463]

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = 2*10^{8}  N/m^{2}

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/(2*10^{8}) = 10^{-5} m^2

recall that area = \pi d^{2} /4

10^{-5} = \frac{3.142*d^{2} }{4} = 0.7855d^{2}

d^{2} = \frac{10^{-5} }{0.7855} = 1.273*10^{-5}

d = \sqrt{1.273*10^{-5}  } = 3.57*10^{-3} m = 3.57 mm

<em>maximum diameter of  the steel linkage d = 3.57 mm</em>

4 0
3 years ago
Pick a subjectarea/field/topic that you are interested in. For each of the following Bonham- Carver uses of GIS give an example
Vanyuwa [196]

Answer:

I hope following attachment will help you a lot!

Explanation:

3 0
3 years ago
2.5
spin [16.1K]
222 is the answer your welcome
6 0
3 years ago
The pressure in a water line is 1500 kPa. What is the line pressure in (a) lb/ft2units and (b) lbf/in2(psi) units?
n200080 [17]

Answer:

Part A:

1500\ KPa= 31328.145 \frac{lb}{ft^2}

Part B:

1500 KPa=217.55656 \frac{lb}{in^2}(Psi)

Explanation:

Part A:

Line Pressure is 1500 KPa

We need a conversion factor which converts KPa to lb/ft^2.

20.88543 \frac{lb}{ft^2}= 1\ KPa

In order to convert 1500 KPa to lb/ft^2, we proceed as:

1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}

1500 KPa is 31328.145 lb/ft^2

Part B:

We will use the same procedure we did in Part A:

1 ft= 12 in

(1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2

Converting 1500 KPa\  into\  \frac{lb}{in^2}

1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}

1500 KPa=217.55656 \frac{lb}{in^2}(Psi)

1500 KPa is 217.55656 lb/in^2 (psi)

3 0
3 years ago
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