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: The total vehicle delay is
39sec/veh
Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.
Effective red time (r) = 25sec
Arrival rate (A) = 900veh/h = 0.25veh/sec
Departure rate (D) = 1800veh/h = 0.5veh/sec
STEP1: FIND THE TRAFFIC INTENSITY (p)
p = A ÷ D
p = 0.25 ÷ 0.5 = 0.5
STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE
The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.
Dt = (A × r^2) ÷ 2(1 - p)
Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)
Dt = 156.25 ÷ 4 = 39.0625
Therefore the total vehicle delay after one cycle is;
Dt = 39
Answer:
speed by mass attain is 55.86 m/s
Explanation:
given data
glucose = 10 g
mass = 100 kg
to find out
speed by mass attain
solution
we know glucose have 180 g molecular weight and
that 1 g glucose produce energy = 2816/180 × 10³ J
so here 10 g of glucose produce energy = 1.56 ×
J
so here energy release = 1/2 × mv²
1.56 ×
= 1/2 × (100)v²
v² = 3.12 × 10³
and v = 55.86 m/s
so speed by mass attain is 55.86 m/s
Answer:
True
Explanation:
This is a true fact because From the time you are born to around the time you turn 30, your muscles grow larger and stronger.
Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.