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3 years ago

A designer chooses a font size and color for a web page. What language was most likely to have been used?

1 answer:

5 0

The language was most likely to have been used is HTML as it help the designer to choose a font size and color for a web page.

<h3>What is HTML a programming language? </h3>

HTML is known as HyperText Markup Language. This is known to be a computer code that is often employed to put together or structure a web page and all of its content.

Note that The language was most likely to have been used is HTML as it help the designer to choose a font size and color for a web page Because HTML is regarded as the markup language that is often used to edit web pages.

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Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been d

navik [9.2K]

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

7 0

3 years ago

Read 2 more answers

Many radios can be operated either by plugging them into the wall or by using batteries. How can a radio use either source of cu

ch4aika [34]

Answer:

The electric current from the batteries installed in a radio supplies direct current (DC) electricity to the radio components directly as an alternative source to the Alternating Current (AC) converted to DC by the power unit located at the radio end of the cable plugged into the wall outlet.

Explanation:

Part of the power unit in a radio includes an AC to DC converter, which is an electrical circuit that is able to convert the alternating current power input from the wall outlet into a direct current output to the radio with which the radio can work

The alternative source of electric current from the batteries installed in a radio bypasses the AC to DC converter and supplies power directly to the radio so it can also work.

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3 years ago

BRO FNAF ANIMATRONICS ALL OF THEM INCLUDING THE BOOKS AND THE OTHER ONES IN FNAF SPECIAL DELIVERY!

Nezavi [6.7K]

Nice nice. . . . . . .

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3 years ago

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The heat rate is essentially the reciprocal of the thermal efficiency. a)- True b)- False

jeyben [28]

Answer:

a). TRUE

Explanation:

Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.

Mathematically, thermal efficiency is

η = net work done / heat input

While heat rate is the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.

Mathematically, heat rate is

Heat rate = heat input / net work done

Thus from above we can see that heat rate is the reciprocal of thermal efficiency.

Thus, Heat rate is reciprocal of thermal efficiency.

4 0

3 years ago

A screw extruder is 50 mm in diameter, 1 m long, has a 50mm lead, a channel 5 mm deep and a flight 3 mm wide. The circular die t

Margarita [4]

Answer:

A) 105.7 rpm

B) 11.32 kw

C) 20.85 NPa

Explanation:

Number of solid rods to be manufactured = 3600

<u>a) Determine the RPM of the screw</u>

we will apply the relation below

discharge rate ( Qd ) = 0.5 π^2 * D^2 * N di * sinA * cos A ------- ( 1 )

where : D = 50 mm , di = 5 mm , N = ?

Tan A = p / πD = 50 / π*50 ∴ A = 17.65°

Insert values into equation ( 1 )

Qd = 17.83 * 10^-6 * N

required discharge rate ( Q ) = $\frac{\frac{\pi D^2}{4}*L*N }{Time}$ ------ ( 2 )

where : D = 0.01 , L = 25 * 10^-2 , N = 3600 , time = 10 * 3600

input value into 2

Q = 31.415 * 10^-6 m^3/s

<em>Hence the RPM of the screw ( N )</em>

= Q / Qd = 31.415 * 10^-6 / 17.83 * 10^-6 = 1.76 rev/s = 105.7 rpm

<u>b) Determine the power requirements of the extruder</u>

max power requirement = Pm * A * πDN / 60

= ( 20.85 * π * ( 50 )^2 / 4 ) * π * 150 *1.76

max power requirement = 11.32 kw

<u>c) What is the pressure buildup within the extruder</u>

Pressure buildup within the extruder = ( 6π*D*N*L* η * cot A ) / di^2

= ( 6π * 0.05 * 1.76 * 1 * 100 * cot17.65 ) / ( 5 * 10^-3 )^2

therefore ; Pm = 20.85 NPa

3 0

3 years ago