Multiple Choice
2 answers:
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Answer:
T = 20.42 N
Explanation:
given data
standard altitude = 30,000 ft
velocity Ca = 500 mph = 0.4 m/s
inlet areas Aa = 7 ft² = 0.65 m²
exit areas Aj = 4.5 ft² = 0.42 m²
velocity at exit Cj = 1600 ft/s = 487.68 m/s
pressure exit j = 640 lb/ft² = 0.3 bar
solution
we get here thrust of the turbojet that is express as
thrust of the turbojet T = Mg × Cj - Ma × Ca + ( j Aj -
a Ag ) .............1
here Ma = Mg
Ma = a × Ca Aa = 0.042 kg/s
put value in equation 1 we get
T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )
T = 20.42 N
Answer:
V = 30 V
vector (E) = -10*a_p + 17.32*a_Q - 24*a_z
mag(E) = 31.24 V/m
dV / dN = 31.2 V /m
a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z
p_v = 276 pC / m^3
Explanation:
Given:
- The Volt Potential in cylindrical coordinate system is given as:
- The point P is at p = 3 , Q = 60 , z = 2
Find:
values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space
Solution:
a)
Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:
b)
To compute the Electric field from Volt potential we have the following relation:
E = - ∀.V
Where, ∀ is a del function which denoted:
∀ =
Hence,
Plug in the values for point P:
c)
The magnitude of the Electric Field is given by:
E = √((-10)^2 + (17.32)^2 + (24)^2)
E = √975.9824
E = 31.241 N/C
d)
The dV/dN is the Field Strength E in normal to the surface with vector N given by:
dV / dN = | - ∀.V |
dV / dN = | E | = 31.2 N/C
e)
a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:
f)
The charge density in free space:
p_v = E.∈_o
Where, ∈_o is the permittivity of free space = 8.85*10^-12
p_v = 31.24.8.85*10^-12
p_v = 276 pC / m^3