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2 years ago

Why would the shear stress be considered as the momentum flux.

Engineering

1 answer:

oksano4ka [1.4K]2 years ago

3 0

Answer:

A fluid flowing along a flat plate will stick to it at the point of contact

Explanation:

and this is known as the no-slip condition. ... This is the precise reason why shear stress in a fluid can also be interpreted as the flux of momentum.

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What is the acceleration of a car that has a velocity of 20 m/s, and

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Acceleration of Car = 10 ms⁻²

Explanation:

Step 1:

The basic formula of acceleration is a = (v-u)/t ms⁻²

where, v- final velocity

u- initial velocity

t= time taken

Step 2:

Here v = 70 ms⁻¹

u = 50 ms⁻¹

t = 5 s

∴ a = ( 70 - 20)/5

a = 10 ms⁻²

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4 years ago

A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to

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Solution :

Given :

The power of the air‐conditioning (AC) unit is , W = 0.434 kW

The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by = 6.22

Therefore he heat removed is given by , $Q_H = 6.22 \times 0.434$

$Q_H = 2.7 \ kW$

Now if the electricity is valued at 0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24

= 6.48

Therefore the operating cost = $ 6.48 for 24 hours.

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3 years ago

1. The WAB Bank decided to open up two new drive-through teller windows outside its main office building in Sariaya, Quezon. The

yan [13]

Plz give me brainalist answer

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3 years ago

A landowner and a contractor entered into a written contract under which the contractor agreed to build a building and pave an a

Leya [2.2K]

The contractor definitely breached the contract

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3 years ago

Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–

Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

$C_p$ = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = $m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]$

Substitute all parameters in the equation

Δs = $5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

Δs = 5 kg × 0.14629 kJ/kg.K

= 0.73145 kJ/K

Δs = $m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]$

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = $5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

= 5 kg (0.13795 kJ/kg.K)

= 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

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4 years ago