r.a.t.e this car from 1/10

Answer:

8 Tips on Better Communication with the Opposite Sex

Put emotions away. Ladies, this one is more aimed at us, for the most part. ...

Forget your pride. In discussions, especially these days, people always want to be the one that prevails. .

Put yourself in their shoes. .

Listen. ...

Respond. ...

Actually communicate. ...

Be detailed. ...

Don't communicate too much.

Explanation:

Answer:

the first cleaning be scheduled 1.006 years after installation

Explanation:

 Given the data in the question;

U[tex]_{clean[/tex] = 300 W/m².K

first we determine the heat coefficient of the dirt surface;

overall heat transfer coefficient is reduced from its initial value by 25%

U[tex]_{dirt[/tex] = ( 1 - 25%) × U[tex]_{clean[/tex]

U[tex]_{dirt[/tex] = ( 1 - 0.25) × 300

U[tex]_{dirt[/tex] = 0.75 × 300

U[tex]_{dirt[/tex] = 225 W/m².K

next we find the inner fouling factor

[tex]R"_{f ,i[/tex] = [tex]a_it[/tex]

[tex]R"_{f ,o[/tex] = (2.5 × 10⁻¹¹)t

for the outer fouling water;

[tex]R"_{f ,o[/tex] = [tex]a_ot[/tex]

[tex]R"_{f ,o[/tex] = ( 1.0 × 10⁻¹¹ )t

now, we determine the total heat transfer coefficient

[tex]\frac{1}{U}[/tex] = [tex]R"_{f ,i[/tex] + [tex]R"_{f ,o[/tex]

we substitute

[tex]\frac{1}{U}[/tex] =  (3.5 × 10⁻¹¹)t

so the first cleaning duration after insulation will be;

[tex]\frac{1}{U} = \frac{1}{U_{dirt}} - \frac{1}{U{clean}}[/tex]

we substitute

(3.5 × 10⁻¹¹)t = [tex]\frac{1}{225} - \frac{1}{300}[/tex]

(3.5 × 10⁻¹¹)t = 0.001111

t = 0.001111 / (3.5 × 10⁻¹¹)

t = 31742857.142857 seconds

t = 31742857.142857 / 3.154 × 10⁷

t = 1.006 years

Therefore, the first cleaning be scheduled 1.006 years after installation

Answer:

6.27 × 10⁻⁴

Explanation:

Relative roughness, k = ε/D where ε = absolute roughness = 0.0025 mm and D = internal pipe diameter = 0.157 in = 0.157 × 25.4 mm = 3.9878 mm

So, k = ε/D

= 0.0025 mm/3.9878 mm

= 6.27 × 10⁻⁴

The relative pipe roughness for a plastic pipe will be:

"6.27 × 10⁻⁴".

According to the question,

Absolute roughness, ε = 0.0025 mm

Internal pipe diameter, D = 0.157 in or,

                                          = 0.157 × 25.4 mm

                                          = 3.9878 mm

We know that,

The relative roughness be:

→ k = [tex]\frac{Absolute \ roughness}{Diameter}[/tex]

or,

   k = [tex]\frac{\varepsilon }{D}[/tex]

By substituting the above values,

      = [tex]\frac{0.0025}{3.9878}[/tex]

      = 6.27 × 10⁻⁴

Thus the above approach is correct.

Find out more information about diameter will be:

https://brainly.com/question/13100709

Answer:

Weight of block is 191.424 Kg

Explanation:

The volume of rectangular block = [tex]1*0.6*0.4 = 0.24[/tex] cubic meter

1/5th of its volume being out of water which means water of volume nearly 4/5 th of the volume of rectangular block is replaced

Volume of replaced water = [tex]\frac{4}{5} * 0.24 = 0.192[/tex] cubic meter

Weight of replaced water = weight of rectangular block = [tex]0.192 * 997[/tex] Kg/M3

= 191.424 Kg

Answer:

15x

Explanation:

Answer:

PERT Chart and GANTT Chart

As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart.  With the PERT chart, the sequence of tasks is clearly mapped out.  Dependent tasks are carried out when other tasks that they depend on have been executed.

Explanation:

By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars.  On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams.  It displays all the project tasks in separate boxes.  The boxes are then connected with arrows which clearly show the task dependencies.

Answer:

Valvular stenosis , Valvular prolapse , Regurgitation,

Explanation:

Answer:

Answer is A

Explanation:

Doesn't break 360 degrees of bend.

The bend that can be used in conduit run is one 90-degree, three 45-degree, and four 30-degree.

Bend that can be used in conduit run is equal or less than 360 degree.

So, One 90-degree, three 45-degree, and four 30-degree is the bend that can be used in conduit run.

Learn more: brainly.com/question/21199524

Answer:

true

Explanation:

the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once

Answer

a) 62 percent

b) 40 percent

Explanation:

Original diameter ( d[tex]_{i}[/tex] ) = 10.33 mm

Original Gauge length ( L[tex]_{i}[/tex] ) = 52.8 mm

diameter at point of fracture ( d[tex]_{f}[/tex] ) = 6.38 mm

New gauge length ( L[tex]_{f}[/tex] ) = 73.9 mm

Calculate ductility in terms of

a) percent reduction in area

percentage reduction = [ (A[tex]_{i}[/tex] - A[tex]_{f}[/tex] ) / A[tex]_{i}[/tex] ] * 100

A[tex]_{i}[/tex] ( initial area ) = π /4 di^2

= π /4 * ( 10.33 )^2 = 83.81 mm^2

A[tex]_{f}[/tex] ( final area ) = π /4 df^2

= π /4 ( 6.38)^2 = 31.97 mm^2

hence : %reduction = ( 83.81 - 31.97 ) / 83.81

= 0.62 = 62 percent

b ) percent elongation

percentage elongation = ( L[tex]_{f}[/tex] - L[tex]_{i}[/tex] ) / L[tex]_{i}[/tex]

= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent

Answer: i got you its d

Explanation:had the smae question as you

9514 1404 393

Answer:

  1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

Answer:

pelo o que diz na database é que você n é ser humano normal por perguntar isso!!

Answer:

In the absence of dark energy, a flat universe expands forever but at a continually decelerating rate, with expansion asymptotically approaching zero; with dark energy, the expansion rate of the universe initially slows down, due to the effects of gravity, but eventually increases, and the ultimate fate of the universe ...

Explanation:

I think it goes on forever.

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v[tex]_f[/tex] = 0.001057 m³/kg

v[tex]_g[/tex] = 1.0037 m³/kg

u[tex]_f[/tex] = 486.82 kJ/kg

u[tex]_g[/tex] 2524.5 kJ/kg

h[tex]_g[/tex] = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v[tex]_g[/tex]

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m[tex]_{out[/tex] = m₁ - m₂

m[tex]_{out[/tex] = 1.89414  - 0.003985

m[tex]_{out[/tex] = 1.890155 kg

so, Initial internal energy will be;

U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]

U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex]  + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E[tex]_{in[/tex] -  E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]

QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁

QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

Answer:

A)

shear plane angle = 31.98°

shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )

B) shear strength = 7339.78

Explanation:

a) Determine the shear plane angle and shear strain

Given data :

Chip thickness before chip formation = 0.5 inches

Chip thickness after separation = 1.125 inches

rake angle ( ∝ ) = 10°  

shear plane angle : Tan ∅ = [tex]\frac{rcos\alpha }{1-sin\alpha }[/tex]    ----- ( 1 )

r = chip thickness ratio = 0.5 / 1.125 = 0.4444

back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10

Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736  = 0.5296

hence ∅ = tan^-1 ( 0.5296 ) = 31.98°

shear strain :  R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )

R = cot ( 31.98° ) + tan ( 31.98 - 10 )

  B) determine the shear strength of the material

cutting force = 1559 N

thrust force = 1271 N

width of cut ( diameter ) = 3.0 mm

shear strength = c + σ.tan ∅

c = cohesion force  = 1271 * 3  = 3813

σ = normal stress  = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94

hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78

Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

= (101 ft - 100 ft) ÷ (100 ft) × 200 ft

= 2 ft

Now the correct distance is

= 200 ft +  2 ft

= 202 ft

Hence, the correct distance is 202 ft

The same would be relevant and considered too

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ /  12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

[tex]E^z[/tex] = - ( εˣ  / v )

[tex]E^z[/tex] = - ( -0.00088  / 0.34 )

[tex]E^z[/tex] = - ( - 0.002588 )

[tex]E^z[/tex] = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

Answer:

a) Density of the road segment

i) Before bus strike = 23.02 pc/mi/In

ii) after bus strike = 25.76 pc/mi/In

b) Volume to capacity ratio

i) Before bus strike = 0.69

ii) After bus strike = 0.78

C) level of service of the upgrade segment

i) Before bus strike = LOS C

ii) after bus strike = LOS D

Explanation:

a) Density of the road segment

i) before bus strike ( D1 )

 D1 = 1662 / 72.18  = 23.02 pc/mi/In

ii) After bus strike ( D2 )

D1 = 1859 / 72.18 = 25.76 pc/mi/In

b) Volume to capacity ratio

i) Before bus strike ( v 1 )

V1 = 1662 / 2400 = 0.69

ii) After bus strike ( V2 )

V2 = 1859 / 2400 = 0.78

C) level of service of the upgrade segment  ( Gotten from " LOS Criteria for basic freeway segments " )

i) Before bus strike =  ( LOS C )

ii) After bus strike  = LOS D

Attached below is the detailed solution to the question above

Answer:

a) the maximum load is 88,040 N

b)

the maximum length to which the specimen may be stretched is 0.12032148 mm

Explanation:

Given the data in the question;

the stress at which plastic deformation begins σ = 284 MPa = 2.84 × 10⁸ Pa

modulus of elasticity E = 106 GPa = 1.06 × 10¹¹ Pa

a)

Area A = 310 mm² = 310 × 10⁻⁶ m ( without plastic deformation )

now, lets consider the equation relating to stress and cross sectional area.

σ = F / A₀

hence, maximum load F = σA₀  

so we substitute

F = (2.84 × 10⁸) × (310 × 10⁻⁶)

F = 88,040 N

Therefore, the maximum load is 88,040 N

b)

Initial length specimen l₀ = 120 mm  = 120 × 10⁻³ m

using engineering strain, ε = (l₁ - l₀)/l₀

Also from Hooke's law, σ = Eε

so from the equation above;

l₁ = l₀( ε + 1 )

l₁ = l₀( σ/E + 1 )

so we substitute

l₁ = (120 × 10⁻³)( (2.84 × 10⁸)/(1.06 × 10¹¹)) + 1 )

l₁ = (120 × 10⁻³) ( 1.002679 )

l₁ = 0.12032148 mm

Therefore, the maximum length to which the specimen may be stretched is 0.12032148 mm

Answer:

D. Categorizing

Explanation:

pls mark me as your brainlist

Answer:

D

Explanation:

Answer:

true

Explanation:

Answer:

45.32%

Explanation:

Given data:

pressure of high temperature steam = 6 MPa

low temperature heat rejection process ( Tr )  = 300 k

A) plot of cycle in t-s coordinates showing steam dome

attached below

B) Calculate thermal efficiency

thermal efficiency = 1 - (Tr / Tsat )

Tsat = 275.59°C  ≈  548.59 K  ( from steam table at Pa = 6 MPa )

back to equation 1

1 - (300 / 548.59 )

1 - 0.5468 = 0.4532 = 45.32%

Answer:

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation )

Explanation:

Yield strength = 275 Mpa

Tensile strength = 380 Mpa

elastic modulus = 103 GPa

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation ) .

Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given

strain = yield strength / elastic modulus

          = 0.0027

Answer:

max Diameter = 0.530 mm

Explanation:

Calculate the maximum Diameter that the thermocouple should have

applying this formula : e = [tex]\frac{SvCv}{hA}[/tex]  ------ ( 1 )

mass = density * volume

Time constant = mc / hA

attached below is the detailed solution

r ( diameter ) = 0.530 mm

the picture is blank for me what does it say i can comment the answer plz mark brainlyist

Answer:

Which option identifies the step of the implementation phase represented in the following scenario?

A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.

establishing a process and budget

using communication tools

building and assembling a team

setting up a change order process

Explanation:

#carryonml

Answer:

using communication tools

Explanation:

The correct answer is using communication tools. Communication tools such as online surveys help project teams identify customers’ wants and needs.

Answer:

[tex]\mathbf{G_m = 2.25}[/tex]

Explanation:

From the given information:

Let the weight of the mix in the air be = [tex]W_{ma}[/tex]

Let the weight of the mix in water be = [tex]W_{mw}[/tex]; &

the bulk specific gravity be = [tex]G_m[/tex]

SO;

[tex]W_{mw} = W_{ma} - v \delta _{w} --- (1)[/tex]

Also;

[tex]G_m = \dfrac{W_{mw}}{v \delta_w} --- (2)[/tex]

From (2), make[tex]v \delta_w[/tex] the subject:

[tex]v \delta_w = \dfrac{W_{ma}}{G_m}[/tex]

Now, equation (1) can be rewritten as:

[tex]W_{mw} = W_{ma} - \dfrac{W_{ma}}{G_m}[/tex]

[tex]G_m = \dfrac{W_{ma}}{W_{ma} - W_{mw}}[/tex]

Replacing the values;

[tex]G_m = \dfrac{1173.5}{1173.5 -652.5}[/tex]

[tex]G_m = \dfrac{1173.5}{521}[/tex]

[tex]\mathbf{G_m = 2.25}[/tex]

Answer: Both technicians A and B

Explanation:

I took the pf test