Answer:
1) Dimensions of shear rate is ![[T^{-1}]](https://tex.z-dn.net/?f=%5BT%5E%7B-1%7D%5D)
.
2)Dimensions of shear stress are ![[ML^{-1}T^{-2}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D)
Explanation:
Since the dimensions of velocity 'v' are ![[LT^{-1}]](https://tex.z-dn.net/?f=%5BLT%5E%7B-1%7D%5D)
and the dimensions of distance 'y' are ![[L]](https://tex.z-dn.net/?f=%5BL%5D)
, thus the dimensions of 
become
![\frac{[LT^{-1}]}{[L]}=[T^{-1}]](https://tex.z-dn.net/?f=%5Cfrac%7B%5BLT%5E%7B-1%7D%5D%7D%7B%5BL%5D%7D%3D%5BT%5E%7B-1%7D%5D)
and hence the units become 
.
Now we know that the dimensions of coefficient of dynamic viscosity 
are ![[ML^{-1}T^{-1}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D)
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}]](https://tex.z-dn.net/?f=%5B%5Ctau%20%5D%3D%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D%5Ctimes%20%5BT%5E%7B-1%7D%5D%5C%5C%5C%5C%5B%5Ctau%20%5D%3D%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D)
Now we know that dimensions of momentum are ![[MLT^{-1}]](https://tex.z-dn.net/?f=%5BMLT%5E%7B-1%7D%5D)
The dimensions of 
are ![[L^{2}T]](https://tex.z-dn.net/?f=%5BL%5E%7B2%7DT%5D)
Thus the dimensions of ![\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]](https://tex.z-dn.net/?f=%5Cfrac%7BMoumentum%7D%7BArea%5Ctimes%20time%7D%3D%5Cfrac%7BMLT%5E%7B-1%7D%7D%7BL%5E%7B2%7DT%7D%3D%5BMLT%5E%7B-2%7D%5D)
Which is same as that of shear stress. Hence proved.
Answer:
phase measurement and the information content
Explanation:
The full form of RTK is Real Time Kinematic. It is used for satellite navigation technique to increase the precision of the position data that is derived from the positioning systems based on satellites like the NavIC, GPS, Galileo, BeiDou and GLONASS. It takes help of the measurements of phase of signal's carrier wave and also the information content of these signals and it also relies on the single interpolated virtual station in order to provide the real time corrections and provide correct and accurate information.
Answer:
COP = 0.090
Explanation:
The general formula for COP is:
COP = Desired Output/Required Input
Here,
Desired Output = Heat removed from water while cooling
Desired Output = (Specific Heat of Water)(Mass of Water)(Change in Temperature)/Time
Desired Output = [(4180 J/kg.k)(3.1 kg)(25 - 11)k]/[(12 hr)(3600 sec/hr)]
Desired Output = 4.199 W
And the required input can be given as electrical power:
Required Input = Electrical Power = (Current)(Voltage)
Required Input = (2.9 A)(16 V) = 46.4 W
Therefore:
COP = 4.199 W/46.4 W
<u>COP = 0.090</u>
Answer:
hi-he = 0
pi-pe = positive
ui-ue = negative
ti-te = negative
Explanation:
we know that fir the sub cool liquid water is
dQ = Tds = du + pdv ............1
and Tds = dh - v dP .............2
so now for process of throhling is irreversible when v is constant
then heat transfer is = 0 in irreversible process
so ds > 0
so here by equation 1 we can say
ds > 0
dv = 0 as v is constant
so that Tds = du .................3
and du > 0
ue - ui > 0
and
now by the equation 2 throttling process
here enthalpy is constant
so dh = 0
and Tds = -vdP
so ds > 0
so that -vdP > 0
as here v is constant
so -dP =P1- P2
so P1-P2 > 0
so pressure is decrease here
Answer:
1790 μrad.
Explanation:
Young's modulus, E is given as 10000 ksi,
μ is given as 0.33,
Inside diameter, d = 54 in,
Thickness, t = 1 in,
Pressure, p = 794 psi = 0.794 ksi
To determine shear strain, longitudinal strain and circumferential strain will be evaluated,
Longitudinal strain, eL = (pd/4tE)(1 - 2μ)
eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)
eL = 3.64 x 10-⁴ radians
Circumferential strain , eH = (pd/4tE)(2-μ)
eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)
eH = 1.79 x 10-³ radians
The maximum shear strain is 1790 μrad.
