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Ilya [14]
2 years ago
13

Which company introduced the Windows operating system correct answer plz ​

Engineering
1 answer:
grigory [225]2 years ago
3 0

Answer:

I think Microsoft Corporation

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A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti
natima [27]

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0
2 years ago
You work for a printing company, and you realize that your colleague sent incorrect price quotes to a client. You begin to write
xxTIMURxx [149]

Answer:

The sentence excerpted from the e-mail uses passive voice.  

Given the  purpose of your message, this voice is appropriate.

Explanation:

Because the objective is to remedy the situation a passive voice is great because it emphasizes the action and the object instead of the subject.

We want to emphasize the document and the incorrect information, not our colleague.

4 0
3 years ago
The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be
lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0
2 years ago
A compressible clay layer has a thickness of 3.8 m. After 1.5 yr, when the clay is 50% consolidated, 7.3 cm of settlement has oc
grigory [225]

The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

<h3>How to determine the amount of settlement?</h3>

For a layer of 3.8 m thickness, we were given the following parameters:

U = 50% = 0.5.

Sc = 7.3 cm.

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is: t_{50} = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

t_{50} = 1.5 × (38/3.8)²

t_{50} = 150 years.

For the thickness of 38 m, U₂ is given by:

\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25  \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05

The new settlement after 1.5 year is:

Sc = U₂Sf

Sc = 0.05 × 146

Sc = 7.3 cm.

For time, t₂ = 5 year:

U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25  \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09

The new settlement after 5 year is:

Sc = U₂Sf

Sc = 0.09 × 146

Sc = 13.14 cm.

Read more on clay layer here: brainly.com/question/22238205

8 0
2 years ago
A vertical piston-cylinder assembly contains water. The piston has a mass of 2 kg and a diameter of 7 cm . Assume atmospheric pr
Step2247 [10]

Answer:

for a) F= 744.97 N

for a) F= 167.85 N

for a) F= 764.57 N

Explanation:

the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.

Then from steam tables

T= 120°C → P required=Pr= 198.67 kPa

then the pressure developed by the piston is

P = (m*g + F)/A

where m= mass of the piston ,g= gravity F= force required and A= area of the piston

then

Pr = P = (m*g + F)/A

F = Pr*A-m*g

since A= π/4*D²

F =π/4* Pr*D²-m*g

replacing values

F =π/4* Pr*D²-m*g = π/4*198.67  *10³Pa*(0.07m)² -2kg* 9.8m/s²

F= 744.97 N

b) for T₂=80°C → Pr₂=47.41 kPa

F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²

F₂= 167.85 N

c) for m=0 (mass of the piston neglected) ,the force required is

F₃ =π/4*Pr*D² =  π/4*198.67 *10³Pa*(0.07m)²= 764.57 N

F₃ =764.57 N

4 0
3 years ago
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