Answer: 15%
Explanation:
Frequency of obtaining = 63/420 = 0.15 = 15%
Answer:
water sample have more water content
Explanation:
given data
soil 1 is saturated with water
unit weight of water = 1 g/cm³
soil 2 is saturated with alcohol
unit weight of alcohol = 0.8 g/cm³
solution
we get here water content that is express as
water content =
....................1
here soil is full saturated so
is 100% in both case
so put here value for water
water content = 100 % × 1
water content = 1 g
and
now we get for alcohol that is
water content = 100 % × 0.8
water content = 0.8 g
so here water sample have more water content
Answer:
Go to explaination for the details of the answer.
Explanation:
Angle of friction (friction angle) is a measure of the ability of a unit of rock or soil to withstand a shear stress. It is the angle (φ), measured between the normal force (N) and resultant force (R), that is attained when failure just occurs in response to a shearing stress (S).
The cohesion intercept is a used when we want to describe the shear strength soils. The definition is mainly derived from the Mohr-Coulomb failure criterion and it is used to describe the non-frictional part of the shear resistance which is independent of the normal stress.
Please check attachment to Determine the friction angle and cohesion intercept.
Answer:
- 1.55
- 260 N.s
- 3370 m
- 1.6
- 43.75 kg/s
Explanation:
1) Thrust coefficient at sea level.
Cfsl = TSL / Pca
TSL = Mp * Vc + ( Pc - Pa )Ac
Mp = mass flux = 43.75 kg/s
∴ Cfsl = Mp Vc / Pca + ( Pc - Pa )/Pc * ( Ac / A* )
= 1.6 - 0.04923 = 1.55
<u>2) Specific impulse at sea </u>
Isp = Vc / g = 2549.75 / 9.81
= 260 N.s
3) Altitude at optimal expansion
H = 3370 m
<u>4) thrust coefficient at optimal expansion </u>
CF = 1.6
attached below is the detailed solution
<u>5) Mass flux through the throat </u>
Mass flux = P1 * At / Cc
= ( 7*10^6 * 0.01 ) / 1600
= 43.75 kg/s
Answer:
(a) 6.91 mm (b) 160 MPa
Explanation:
Solution
Given that:
E = 200 GPa
The rod length = 48 mm
P =P¹ = 6 kN
Recall that,
1 kN = 10^3 N
1 m =10^3 mm
I GPa = 10^9 N/m²
Thus
The rod deformation is stated as follows:
δ = PL/AE-------(1)
σ = P/A----------(2)
Now,
(a) We substitute the values in equation and obtain the following:
48 * 10 ^⁻3 m = (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]
Thus, we simplify
A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²
A =0.0375 * 10 ^⁻3 m²
A =37.5 mm²
A = π/4 d²
Thus,
d² = 4A /π
After inserting the values we have,
d = √37.5 * 4/3.14 mm
= 6.9116 mm
or d = 6.91 mm
Therefore, the smallest that should be used is 6.91 mm
(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)
Thus,
σ = P/A
σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²
σ= 160 MPa
Note: I MPa = 10^6 N/m²
Hence the the corresponding normal stress is σ= 160 MPa