Answer:
HUMAN DEVELOPMENT
MOTOR BEHAVIOR
EXERCISE SCIENCE
MEASUREMENT AND EVALUATION
HISTORY AND PHILOSOPHY
UNIQUE ATTRIBUTES OF LEARNERS
CURRICULUM THEORY AND DEVELOPMENT
Explanation:
Answer:
true
Explanation:
it depends on how fast the car goes because the gas burns faster or slower depending on the speed you go or drive. if your drive 50 MPH, the gas will burn slowly. if your drive 45 MPH, the gas will burn faster if you go slow too much. if you press the glass button and hold it for 5 minutes, the more you hold the gas for too long, the more the gas will burn inside the gas engine. go look at your car or your parents car and see how it goes and that will help.
Answer:
Explanation:
(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄
Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14
1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54
= 153.12 + 21.28 + 148.16 + 7.56
= 330.12 g/mol
which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable
The moles of methane produced will be given as
m = (4a + b -2c - 3d)/8
= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8
= (51.04 + 21.28 - 18.52 - 1.62)/8
= 52.18/8
= 6.5225
(b) Volume of methane V is given as
V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)
= 0.1461 m³ CH₄/kg lawn trimmings
(c) Energy will be given as
CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol
= 5805.025
≈ 5805 kJ/kg
Answer:
mean:24/5
ascending order:1,2,3,4,5,5,6,6,7,9
median:5
Answer:
Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56 A
Explanation:
In physics, the determination of the term " natural response" simply means that we want to know what happens in a circuit when the value of t = 0, that is to say after the circuit has been disconnected. Hence, the value of the voltage and the current can then be determined or Calculated;
For the the natural response of i in(t) we will be using the formula below;
I(t) = Vo × t/ R = Vo/R × e^-t/h.
Where h = 1/RC = time constant.
For t= 0^- = 0.8 × 1= 0.8 V.
1/Ctotal = 1/ 2 + 1/3 = 6/5
For t = 0^+;
h =( 0.8 + 0.5) × 6/5 = 1.56 seconds.
Hence, we will have;
Vin(t) = 0.8 × e^-t/1.56.
Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56
