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What type of companies would employ in mechanics engineering

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Alex73 [517]3 years ago

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What do y’all do when ya girl go eat lunch and eat it and eat

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What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ

swat32

Answer:

1) Dimensions of shear rate is $[T^{-1}]$ .

2)Dimensions of shear stress are $[ML^{-1}T^{-2}]$

Explanation:

Since the dimensions of velocity 'v' are $[LT^{-1}]$ and the dimensions of distance 'y' are $[L]$ , thus the dimensions of $\frac{dv}{dy}$ become

$\frac{[LT^{-1}]}{[L]}=[T^{-1}]$ and hence the units become $s^{-1}$ .

Now we know that the dimensions of coefficient of dynamic viscosity $\mu$ are $[ML^{-1}T^{-1}]$ thus the dimensions of shear stress can be obtained from the given formula as

$[\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}]$

Now we know that dimensions of momentum are $[MLT^{-1}]$

The dimensions of $Area\times time$ are $[L^{2}T]$

Thus the dimensions of $\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]$

Which is same as that of shear stress. Hence proved.

4 0

3 years ago

A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank

Bad White [126]

Given:

mass of water, m = 2000 kg

temperature, T = $30^{\circ}C$ = 303 K

extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m $\bar{R}$ T (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

$V = \frac{m\bar{R}T}{P}$

$V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}$

$V = 2762.44 m^{3}$

Therefore, the volume of the tank is $V = 2762.44 m^{3}$

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = $30^{\circ}C$ = 303 K

At equilibrium, volume remain same

So,

P'V = m' $\bar{R}$ T'

$P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}$

Therefore, the final pressure is P' = 96.258 kPa

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Only put coolant into your radiator when the engine is _____.

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A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

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