Answer:
The additive drag at flight condition will be found by the following equation
Area = A1 = 6m2
Da = Additive drag
Cda = Additive drag coefficient
P = prassure at altitude of 12Km
Po = Prassure at sea level
\gamma = Ratio of specific heat capacity
The formula of additive drag is given below
D = Cda q A
q = Dynamic prassure , A = cross sectional area
q =( \gamma/ 2) P0 M02
D = Cda ( \gamma/2) po Mo2 A
Cda=0.32
\gamma=1.4
M=0.8
p0= 101325pa
D = 0.32 (1.4/2)(101325pa)(0.6)2 6
D = 49025N
Explanation:
The additive drag at the flight conditions will be D= 49025N
Answer:
For a Singular matrix, the determinant must be equivalent to 0.
Explanation:
A matrix is a rectangular array in which elements are arranged in rows and columns.
Each square matrix has a determinant. The determinant is a numerical idea that has a fundamental function in finding the arrangement just as investigation of direct conditions. For a Singular matrix, the determinant must be equivalent to 0.
Answer:
P_p = 27000 psf
Explanation:
given,
height of the retaining wall = h = 12 ft
internal angle of friction (∅)= 30°
unit weight = 125 pcf
Rankine passive earth pressure = ?
k_p is the coefficient of passive earth pressure


k_p = 3
Passive earth pressure


P_p = 27000 psf
Rankine passive earth pressure on the wall is equal to P_p = 27000 psf
He started work at 16 for the North West Company and then the Hudson's Bay Company, becoming a high-ranking officer. From 1851 to 1864, he was Governor of the Colony of Vancouver Island
Answer:
2.77mpa
Explanation:
compressive strength = 20 MPa. We are to find the estimated flexure strength
We calculate the estimated flexural strength R as
R = 0.62√fc
Where fc is the compressive strength and it is in Mpa
When we substitute 20 for gc
Flexure strength is
0.62x√20
= 0.62x4.472
= 2.77Mpa
The estimated flexure strength is therefore 2.77Mpa