Answer:
(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift
Explanation:
When a fluid flows around the surface of an object, it exerts a force on it. This force has two components, namely lift and drag.
The component of this force that is perpendicular (normal) to the freestream velocity is known as lift, while the component of this force that is parallel or in the direction of the fluid freestream flow is known as drag.
Lift is as a result of pressure differences, while drag results from forces due to pressure distributions over the object surface, and forces due to skin friction or viscous force.
Thus, drag results from the combination of pressure and viscous forces while lift results only from the<em> pressure differences</em> (not pressure forces as was used in option D).
The only correct option left is "A"
(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift
Answer:
8 to 10 times
Explanation:
For dry road
u= 15 mph ( 1 mph = 0.44 m/s)
u= 6.7 m/s
Let take coefficient of friction( μ) of dry road is 0.7
So the de acceleration a = μ g
a= 0.7 x 10 m/s ² ( g=10 m/s ²)
a= 7 m/s ²
We know that
v= u - a t
Final speed ,v=0
0 = 6.7 - 7 x t
t= 0.95 s
For snow road
μ = 0.4
de acceleration a = μ g
a = 0.4 x 10 = 4 m/s ²
u= 30 mph= 13.41 m/s
v= u - a t
Final speed ,v=0
0 = 30 - 4 x t'
t'=7.5 s
t'=7.8 t
We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.
8 to 10 times
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-
a=
b=
from virial equation
And
a=
b=
Now calculate V1 and V2 at given condition
Substitute given values 
= 1 x 10^5 , T = 373.15 and given values of coefficients we get
Solve for V1 by iterative or alternative cubic equation solver we get
Similarly solve for state 2 at P2 = 50 bar we get
Now
a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is
a=
b=
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series
