16. Which polymers are recyclable?

Explain why slopping roofs are used in areas with heavy rain or snow fall

Savatey [412]

Answer:

Slopping or slanting roofs are found in houses where there are snow or rain, because it helps the water or the snow to skid down easily. It minimizes the chances of waterlogging because water cannot rest on the roof surface for long.

Explanation:

Brainliest pls.

8 0

3 years ago

In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh

astraxan [27]

( 12 17 18 19 25 )

Explanation:

First Pass:

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

Second Pass:

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

Third Pass:

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

Fourth Pass:

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

Fifth Pass:

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

6 0

3 years ago

Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a g

brilliants [131]

Answer:

The number of germanium atoms per cubic centimeter for this germanium-silicon alloy is 3.16 x 10²¹ atoms/cm³.

Explanation:

Concentration of Ge ( $C_{Ge}$ ) = 15%

Concentration of Si (C $_{Si}$ ) = 85%

Density of Germanium (ρ $_{Ge}$ ) = 5.32 g/cm³

Density of Silicon (ρ $_{Si}$ ) = 2.33 g/cm³

Atomic mass of Ge (A $_{Ge}$ )= 72.64 g/mol

To calculate the number of Ge atoms per cubic centimeter for the alloy, we will use the formula:

No of Ge atoms/cm³=[Avogadro's Number* $C_{Ge}$ ]/([ $C_{Ge}$ *A $_{Ge}$ /ρ $_{Ge}$ )+(C $_{Si}$ *A $_{Ge}$ /ρ $_{Si}$ )]

= (6.02x10²³ * 15%) / [(15% * 72.64/5.32)+(72.64*85%/2.33)]

= (9.03x10²²)/(2.048+26.499)

= (9.03x10²²)/(28.547)

No of Ge atoms/cm³ = 3.16 x 10²¹ atoms/cm³

3 0

3 years ago

what do you think the author is trying to express to society during his time of period through this story? in the open window

jekas [21]

Vroom vroom move uno rass

3 0

4 years ago

A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not e

mash [69]

Answer:

The steel is a candidate.

Explanation:

Given

P = 27,500 N

d₀ = 10.0 mm = 0.01 m

Δd = 7.5×10 ⁻³ mm (maximum value)

This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is applied; this means that the stress corresponding to this load not exceed the yield strength of the material.

Upon computing the stress

σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²

⇒ σ = 350 MPa

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10 ⁻³ mm), it is necessary to calculate the change in diameter Δd for these four alloys.

Then we use the equation υ = - εx / εz = - (Δd/d₀)/(σ/E)

⇒ υ = - (E*Δd)/(σ*d₀)

Now, solving for ∆d from this expression,

∆d = - υ*σ*d₀/E

For the Aluminum alloy

∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm

0.0165 mm > 7.5×10 ⁻³ mm

Hence, the Aluminum alloy is not a candidate.

For the Brass alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm

0.0118 mm > 7.5×10 ⁻³ mm

Hence, the Brass alloy is not a candidate.

For the Steel alloy

∆d = - (0.3)*(350 MPa)*(10 mm)/(207*10³MPa) = - 0.005 mm

0.005 mm < 7.5×10 ⁻³ mm

Therefore, the steel is a candidate.

For the Titanium alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(107*10³MPa) = - 0.0111 mm

0.0111 mm > 7.5×10 ⁻³ mm

Hence, the Titanium alloy is not a candidate.

7 0

3 years ago