Answer: A. Thermohaline circulation
Explanation:
The thermohaline circulation is a phenomena which involves the movement of the oceanic currents because of the differences that occur in the temperature and salinity of oceanic water. These two factors changes the salinity and density of the water and fluctuating the climatic conditions. Cold water is particularly denser than the warm water. The water with more salinity is also denser over water with less salinity. The deep oceanic water currents brings changes in the density as well as the salinity of water hence, the circulation of water is called as thermohaline circulation. Some part of warm water will evaporate from the surface layer of the ocean this will lead to the flow of warm wind currents and water to the regions of cold regions of the seawater.
On the basis of the above description, A. Thermohaline circulation is the correct option.
Answer:
V2 = 21.44cm^3
Explanation:
Given that: the initial volume of the bubble = 1.3 cm^3
Depth = h = 160m
Where P2 is the atmospheric pressure = Patm
P1 is the pressure at depth 'h'
Density of water = ρ = 10^3kg/m^3
Patm = 1.013×10^5 Pa.
Patm = 101300Pa
g = 9.81m/s^2
P1 = P2+ρgh
P1 = Patm +ρgh
P1 = 1.013×10^5+10^3×9.81×160.
P1 = 101300+1569600
P1 = 1670900 Pa
For an ideal gas law
PV =nRT
P1V1/P2V2 = 1
V2 = ( P1/P2)V1
V2 = (P1/Patm)V1
V2 = ( 1670900 /101300 Pa) × 1.3
V2 = 1670900/101300
V2 = 16.494×1.3
V2 = 21.44cm^3
Answer:
0.46786 W
Explanation:
The solution is in the attached file below
Answer:
Explanation:
Let assume that air behaves ideally. The equation of state of ideal gases is:
Where:
- Pressure, in kPa.
- Volume, in m³.
- Quantity of moles, in kmol.
- Ideal gas constant, in 
.
- Temperature, in K.
Since there is no changes in pressure or the quantity of moles, the following relationship between initial and final volumes and temperatures is built:
The final temperature is:
it would be 'B' because it speeds up reactions
