Answer:
1) Dimensions of shear rate is
.
2)Dimensions of shear stress are
Explanation:
Since the dimensions of velocity 'v' are
and the dimensions of distance 'y' are
, thus the dimensions of
become
and hence the units become
.
Now we know that the dimensions of coefficient of dynamic viscosity
are
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:
Option D. is correct
Explanation:
Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment, he is not able to reach the new network segment from his workstation.
The most problem is that the technician used a straight-through cable.
Option D. is correct.
Answer:
volume = 53.747 m3 = 14198.138 gal
weight = 526652 N = 118396.08 lbf
Explanation:
We know that volume of water

where A' = 61% of A


=1898.015 ft^3



=526652 N

Answer:
••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.