Answer:
See below in the explanation section the Matlab script to solve the problem.
Explanation:
prompt='enter the first weight w1: ';
w1=input(prompt);
wd1=double(w1);
prompt='enter the second weight w2: ';
w2=input(prompt);
wd2=double(w2);
prompt='enter the third weight w3: ';
w3=input(prompt);
wd3=double(w3);
prompt='enter the fourth weight w4: ';
w4=input(prompt);
wd4=double(w4);
prompt='enter the first weight w5: ';
w5=input(prompt);
wd5=double(w5);
x=[wd1 wd2 wd3 wd4 wd5]
format short
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:
winter viscosity grades
Explanation:
The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.
Answer:
Mechanical engineer? Thats my guess I didnt have alot of options sorry if I am wrong
Explanation: