<span>The melting and freezing points of a substance are the same.
Substances melt and freeze at the same temperature. Hope this helps! :)</span>
<span>T(t)=60+140<span>e<span>−0.075t</span></span></span>
<span>T(12)=60+140<span>e<span>−0.075∗12</span></span></span>
<span>T(12)=60+140<span>e<span>−0.9</span></span></span>
<span><span>T(12)=60+140(0.4065696597)
=116.84
So the temperature will be approximately 117 degrees</span></span>
When it travels 3m ,4m and 5m it means 12m is right answer.
Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
Hi! Decomposers (mainly soil bacterium, fungus, or invertebrate)<span /> are categorized as consumers due to the fact that they consume dead organic matter such as plants and animals. They differ from producers (green plants and some bacteria) because they do not produce their own food using photosynthesis or chemosythensis.
Hope this helped!
