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blagie [28]
3 years ago
10

Am i eating ramon nooddles rn

Engineering
2 answers:
Kruka [31]3 years ago
7 0

Answer:

yes

Explanation:

Elden [556K]3 years ago
3 0

Answer:

You are eating ramen

Explanation:

It is shrimp flavor, yee yee

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Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0
3 years ago
As cylinder pressure and heat increase due to an increased load condition, the fuel injection management system must ___________
Temka [501]

possible Answers:

Compensate ⭐⭐⭐⭐⭐

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6 0
2 years ago
Why is it that dislocations play an important role in controlling the mechanical properties of metallic materials, however, they
klio [65]

Answer:

dislocations play an important role in controlling as

Explanation:

As dislocations plays an important role in the ductility, elasticity and plurality of materials

  • The elastic and elastic deflections play a large role in their properties as the metallic materials, because the dislocation of a glass material does not play a major role in their properties.

5 0
3 years ago
PLS HELP!!!
Advocard [28]

Answer:

I would say C

Explanation:

let me know if im right

3 0
2 years ago
Read 2 more answers
A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is imp
Nikolay [14]

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

6 0
3 years ago
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