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.
Your Answer would be A I believe.
Answer:
16-bit wide
Explanation:
In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.
If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:
64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.
In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.