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2 years ago

What are some advantages of making electronic components like transistors increasingly smaller?

Engineering

1 answer:

Stells [14]2 years ago

4 0

Answer:

Dr. Engelbart, who would later help develop the computer mouse and other personal computing technologies, theorized that as electronic circuits were made smaller, their components would get faster, require less power and become cheaper to produce — all at an accelerating pace

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Choose the best compression ratio for an Otto cycle reciprocating engine: a. 5 b. 10 c. 15 d. 20

MArishka [77]

Answer:

b)10

Explanation:

We know that in Otto cycle there are four process .In first process mass of air fuel mixture enters in the cylinder.In second process that intake mass is compresses up to a specified limit.In general compression ratio in Otto cycle is about 9 or 10.In third process burning of fuel takes place where certain amount of work is obtained.Last process is exhaust process in which burning gas escape out from the cylinder.

Best compression ratio in Otto cycle is about 10.Beyond this value of compression ratio the cycle will not proved best work out put. So our option b is right.

6 0

4 years ago

For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep

Alisiya [41]

Answer:

MRR = 1.984

Explanation:

Given that

Depth of cut ,d=0.105 in

Diameter D= 1 in

Speed V= 105 sfpm

feed f= 0.015 ipr

Now the metal removal rate given as

MRR= 12 f V d

d= depth of cut

V= Speed

f=Feed

MRR= Metal removal rate

By putting the values

MRR= 12 f V d

MRR = 12 x 0.015 x 105 x 0.105

MRR = 1.984

Therefore answer is -

1.944

8 0

3 years ago

A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observe

Alenkinab [10]

Answer:

moment of inertia = 4.662 * 10^6 $mm^4$

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 $\pi f$ = 58.434 rad/secs

also Wm = $\sqrt{\frac{g}{t} }$ ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = $\frac{wl^3}{48EI}$

insert the value of t into equation 1

W $m^2$ = $\frac{g*48*E*I}{WL^3}$ make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = $\frac{WL^3* Wn^2}{48*g*E}$

I = $\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}$ = 4.662 * 10^6 $mm^4$

6 0

3 years ago

2=333=3= im single text in comment

sasho [114]

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6 0

3 years ago

Read 2 more answers

A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re

Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

4 0

3 years ago