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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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brilliants [131]

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3 years ago

What happens to the amperage draw of a condensing unit on a split AC system if the liquid line is restricted

cupoosta [38]

Answer:

The amperage draw of the condensing unit will be low.

Explanation:

A condensing unit is made up of a compressor and condenser, while an evaporating unit is made up of an evaporator coil.

A split AC system is a type of air conditioner system that has a condensing unit which is placed separately from the evaporative coil unit. Then the two units are connected to each other via a copper tube containing refrigerants.

The liquid line connects the condenser to the evaporator, and if this liquid line is restricted, the amp consumed by the condensing unit will be low.

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3 years ago

"geophysical exploration definition"

Strike441 [17]

Answer:

Exploration geophysics is an applied branch of geophysics and economic geology, which uses physical methods, such as seismic, gravitational, magnetic, electrical and electromagnetic at the surface of the Earth to measure the physical properties of the subsurface, along with the anomalies in those properties

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3 years ago

Read 2 more answers

Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3

Paladinen [302]

Answer:

A) n = 0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min

= - N = $\frac{In(V2) - In(V1)}{In(T2)-ln(T1)}$

therefore - N = $\frac{5.043 - 4.605}{2.302 -3.015}$

= - 0.6143

THEREFORE ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence C ≈ 640m/min

B) Calculate the values of N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n ---------------- equation 2

also using the second values of v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

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3 years ago

Can you determine the critical distance along a flat surface?

Keith_Richards [23]

Explanation:

Consider a fluid of density, ρ moving with a velocity, U over a flat plate of length, L.

Let the Kinematic viscosity of the fluid be ν.

Let the flow over the fluid be laminar for a distance x from the leading edge.

Now this distance is called the critical distance.

Therefore, for a laminar flow, the critical distance can be defined as the distance from the leading edge of the plate where the Reynolds number is equal to 5 x $10^{5}$

And Reynolds number is a dimensionless number which determines whether a flow is laminar or turbulent.

Mathematically, we can write,

Re = $\frac{\rho .U.x}{\mu }$

or 5 x $10^{5}$ = $\frac{\rho .U.x}{\mu }$ ( for a laminar flow )

Therefore, critical distance

$x=\frac{5\times 10^{5}\times \mu }{\rho \times U}$

So x is defined as the critical distance upto which the flow is laminar.

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3 years ago