Answer:
cultivation - preparing and planting crops
domestication - capturing, taming, and breeding animals
hunting and gathering - obtaining food from the wild
Explanation:
moo
Answer:
Macronutrients are simply nutrients the body needs in a very high amount e.g Carbohydrate.
MicroNutrients are simply nutrients the body needs but in little amount e.g Minerals.
Explanation:
So for further breakdown:
What are nutrients? Nutrients are essential elements that nourish the body in different capacities. We as humans get most of out nutrients from the food and water we ingest.
Now about Macro Nutrients: From the prefix "Macro" which means large, we can infer that macro nutrients are elements need by the body for the fundamental processes of the body, deficiency in this nutrients are very easy to spot. Examples are: Carbohydrates, Protein, Fats amd Water.
Micro Nutrients: In relation to macro nutrients this are elements that the body needs but are not needed in Large quantities. They mostly work like supporting nutrients. Most chemical activities like reaction that occur in the body are a function of micro nutrients. Defiencies in micrp nutrients may take some time to spot e.g Minerals and Vitamins
In regards to exercise: Macro nutrients are the essential ones here since they are the ones that generate energy. PS: micro nutrients dont generate energy.
In regards to rest: Both the Macro and Micro Nutrients are essentail for the overall well being of the body.
<u></u>
has greater effect.
<u>Explanation</u>:
= Temperature of cold reservoir
= Temperature of hot reservoir
when 
is decreased by 't',
= 
when 
is increased by 'T'
Answer:
t = 25.10 sec
Explanation:
we know that Avrami equation
here Y is percentage of completion of reaction = 50%
t is duration of reaction = 146 sec
so,
taking natural log on both side
ln(0.5) = -k(306.6)
for 86 % completion
t = 25.10 sec
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
