Answer:
The average velocity is 0.203 m/s
Explanation:
Given;
initial displacement, x₁ = 20 yards = 18.288 m
final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m
change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s
The average velocity is given by;
V = change in displacement / change in time
V = (x₂ - x₁) / Δt
V = (18.288 - 6.096) / 60
V = 0.203 m/s
Therefore, the average velocity is 0.203 m/s
Answer:
F₁ = 1500 N
F₂ = 750 N
= 500 N
Explanation:
Given :
Power transmission, P = 7.5 kW
= 7.5 x 1000 W
= 7500 W
Belt velocity, V = 10 m/s
F₁ = 2 F₂
Now we know from power transmission equation
P = ( F₁ - F₂ ) x V
7500 = ( F₁ - F₂ ) x 10
750 = F₁ - F₂
750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )
∴F₂ = 750 N
Now F₁ = 2 F₂
F₁ = 2 x F₂
F₁ = 2 x 750
F₁ = 1500 N , this is the maximum force.
Therefore we know,
= 3 x 
where
is centrifugal force
=
/ 3
= 1500 / 3
= 500 N
Answer:
(b)False
Explanation:
Given:
Prandtl number(Pr) =1000.
We know that 
Where
is the molecular diffusivity of momentum
is the molecular diffusivity of heat.
Prandtl number(Pr) can also be defined as

Where
is the hydrodynamic boundary layer thickness and
is the thermal boundary layer thickness.
So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
So hydrodynamic layer will be thicker than the thermal boundary layer.
Answer:
a)
, b)
,
,
, c)
,
,
, 
Explanation:
a) The total number of users that can be accomodated in the system is:


b) The length of the side of each cell is:


Minimum time for traversing a cell is:



The maximum time for traversing a cell is:


The approximate time is giving by the average of minimum and maximum times:


c) The total number of users that can be accomodated in the system is:


The length of each side of the cell is:


Minimum time for traversing a cell is:



The maximum time for traversing a cell is:


The approximate time is giving by the average of minimum and maximum times:

