Answer:
Farmers who use conventional tillage use compost more than other farmers.
Explanation:
Convectional tillage is good to the environment because it attains the following;
- It increases porosity of the soil
- It loosens the soil thus allowing proper root growth and air exchange
- It is an effective way to incorporate manure and break sod fields
- Tilled soils warm faster in spring
Answer: speaks Portuguese
Eu disse a todos a tradução para que possam te ajudar
Explanation: Y’all can help I have no idea
QUESTION 13. Explain the use of quotation marks in the excerpt "I will give each of you
seed. The one who will bring me the most beautiful flower within six months will be chosen but
wife and the future empress of China. ".
QUESTION 14. The palace servant considered the idea of her daughter attending the celeb
organized by the prince of the region, a foolish idea, a madness. This is an OP
Do you agree with this opinion of the character? Justify your answer.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) The required additional minterms for f so that f has eight primary implicants with two literals and no other prime implicant are
and 
b) The essential prime implicant are
and 
c) The minimum sum-of-product expression for f are
Explanation:
The explanation is shown on the second third and fourth image
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

For ideal gas,
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.