Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.
<h3>What is the mass of casein in wet casein?</h3>
The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg
Mass of water 250 kg
The mass of casein is constant while the moisture content can be changed.
At 12% moisture content;
750 kg = 88%%
100 % = 100 ×750/88 = 852.27 kg
Therefore, the mass of dried casein produced os 852.3 kg.
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Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
Answer:
Indicators for ineffective system engineering are as follows
1.Requirement trends
2.System definition change backlog trends
3.interface trends
4.Requirement validation trends
5.Requirement verification trends
6.Work product approval trends
7.Review action closure trends
8.Risk exposure trends
9.Risk handling trends
10.Technology maturity trends
11.Technical measurement trends
12.System engineering skills trends
13.Process compliance trends
Answer:
See explanations for step by step procedures to get answer.
Explanation:
Given that;
Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III, and M0M0M_0. Enter positive value if the deflection is upward and negative value if the deflection is downward.
