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german
2 years ago
15

Aqueous cleaners are ________ parts cleaning agents.

Engineering
1 answer:
GREYUIT [131]2 years ago
4 0

Answer:water based

Explanation:

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The pressure of a gas in a rigid container is 125kpa at 300k, what we be the new pressure if the temperature increases to 900k​
kipiarov [429]

Answer:

375 KPa

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 125 KPa

Initial temperature (T₁) = 300 K

Final temperature (T₂) = 900 K

Final pressure (P₂) =?

The new (i.e final) pressure of the gas can be obtained as follow:

P₁/T₁ = P₂/T₂

125 / 300 = P₂ / 900

Cross multiply

300 × P₂ = 125 × 900

300 × P₂ = 112500

Divide both side by 300

P₂ = 112500 / 300

P₂ = 375 KPa

Thus, the new pressure of the gas is 375 KPa

7 0
3 years ago
What is the weight density of a cylinder that has a diameter of 4.0 inches and a height of 10.0 inches? The cylinder weighs 2.0
zhannawk [14.2K]

Answer:

20

Explanation:

4.0 has the diameter of 2.0 lbs so dived it is 20

7 0
4 years ago
A one-dimensional slab without heat generation has a thickness of 20 mm with surfaces main- tained at temperatures of 275 K and
vlada-n [284]

Answer:

a) 512.5 KW/m2

b) 40.75 KW/m2

c) 2 KW/m2

Explanation:

Given data;

T_2 = 325 K

T_1 = 275 K

dx = 0.20 mm

a) for aluminium   K = 205 W/m k

heat flux = k \frac{dt}{dx}

               = 205 \frac{325 - 275}{0.02}

               = 512.5 KW/m2

b) for AISI 316 stainless steel

k = 16.3 W/ m k

heat flux = k \frac{dt}{dx}

               = 16.3 \frac{325 - 275}{0.02}

               = 40.75 KW/m2

C) for Concrete

k = 0.8 W/ m k

heat flux = k \frac{dt}{dx}

               = 0.8 \times \frac{325 - 275}{0.02}

               = 2 KW/m2

6 0
4 years ago
A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T
Maru [420]

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

<h3>How to find the net radiation heat transfer?</h3>

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

Read more about Net Radiation Heat Transfer at; brainly.com/question/14148915

#SPJ1

8 0
2 years ago
Select the parameters that are included in a baseline performance check. you may select more than one. select one or more: a. co
pishuonlain [190]

Answer:

A F E C D B

Explanation:

6 0
3 years ago
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