Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
<u>Calculate the value of the M2 </u>
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
<em>Note: Calculation for T2 is attached below</em>
Answer: preventive maintenance
Explanation:
The modern hydraulic lifts make use of biodegradable fluid to transmit hydraulic power
<em>Question: The options are left out in the question. The details and facts about the modern hydraulic lift are presented here</em>
<em />
Details about the modern hydraulic lifts include;
The development of the modern hydraulic occurred in the Industrial Revolution to perform task done previously by steam powered elevators
The power of the hydraulic lift come from the hydraulic cylinder known as the actuator, which in turn is powered by pressurized hydraulic fluid such as oil
The hydraulic fluid is pushed by a piston rod through which energy is capable of being transferred, such that the applied force is multiplied, to provide more power for lifting
<u>Facts about the modern hydraulic lifts include;</u>
- The dry motor in the modern hydraulic lift is more efficient and consumes 20% less energy
- It comprises of valves that are controlled electronically such that the response is much rapid and the energy consumption is reduced by a further 20%
- The cars used in the modern lift are lighter, as well as the slings, which reduces the power usage by 20%
- It makes use of chemicals which are environmentally friendly as hydraulic fluid
- The flash point of the fluid used is higher, as well as it posses 50% lower compressibility as well elasticity
Learn more here:
brainly.com/question/16942803
Answer:
6 houses
Explanation:
because
2hrs=4 houses which means you are cleaning 2houses in one hour
so in 3 hours you will houses because you will clean 2 houses in one hour
I hope this helped you sorry if I am wrong
Answer:
a.)
US Sieve no. % finer (C₅ )
4 100
10 95.61
20 82.98
40 61.50
60 42.08
100 20.19
200 6.3
Pan 0
b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4
c.) Cu = 3.33
d.) Cc = 1
Explanation:
As given ,
US Sieve no. Mass of soil retained (C₂ )
4 0
10 18.5
20 53.2
40 90.5
60 81.8
100 92.2
200 58.5
Pan 26.5
Now,
Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g
⇒ w = 421.2 g
As we know that ,
% Retained = C₃ = C₂×
∴ we get
US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)
4 0 0
10 4.39 4.39
20 12.63 17.02
40 21.48 38.50
60 19.42 57.92
100 21.89 79.81
200 13.89 93.70
Pan 6.30 100
Now,
% finer = C₅ = 100 - C₄
∴ we get
US Sieve no. Cummulative % retained (C₄) % finer (C₅ )
4 0 100
10 4.39 95.61
20 17.02 82.98
40 38.50 61.50
60 57.92 42.08
100 79.81 20.19
200 93.70 6.3
Pan 100 0
The grain-size distribution is :
b.)
From the diagram , we can see that
D10 = 0.12
D30 = 0.22
D60 = 0.12
c.)
Uniformity Coefficient = Cu = 
⇒ Cu = 
d.)
Coefficient of Graduation = Cc = 
⇒ Cc = 
= 1
