Yup, I think you add all of them
The answer is "Cold Front".
A cold front is characterized as the change zone where a cool air mass is supplanting a hotter air mass. Cold fronts for the most part move from northwest to southeast. The air behind a cold front is recognizably colder and drier than the air in front of it.
Answer:
Explanation:
In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.
The equation given by Newton tells us that
In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.
In the case of , we understand that they are constant.
We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.
In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.
In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line
Answer:
a) T2 = 133.5°C
x2 = 0.364
b) Qout = 3959.6 kJ
Explanation:
For this exercise we will make two assumptions: the tank is stationary, the kinetic and potential energy are equal to zero. The second assumption is that there are interactions between the two tanks. The contents of both tanks will be a closed system. The energy balance equals:
ΔEsystem = Ein - Eout
-Qout = ΔUA + ΔUB = (m*(u2-u1))A + (m*(u2-u1))B
The steam properties for the two tanks in the initial state can be found in Table A-4 to table A-6 of Cengel:
P1A = 1000 kPa
T1A = 300°C
v1A = 0.25799 m^3/kg
u1A = 2793.7 kJ/kg
T1B = 150°C
x1 = 0.5
vf = 0.001091 m^3/kg
uf = 631.66 kJ/kg
vg = 0.39248 m^3/kg
ufg = 1927.4 kJ/kg
v1B = vf + x1*vfg = 0.001091 + (0.5*(0.39248-0.001091)) = 0.19679 m^3/kg
u1B = uf + x1*ufg = 631.66 + (0.5*1927.4) = 1595.4 kJ/kg
V = VA + VB = mA*v1A + mB*v1B = (2*0.25799) + (3*0.19679) = 1.106 m^3
m = mA + mB = 3+2 = 5 kg
the specific volume will be equal to:
v2 = V/m = 1.106/5 = 0.2213 m^3/kg
With these calculations, we can looking the new properties in the same tables:
P2 = 300 kPa
v2 = 0.2213 m^3/kg
T2 = Tsat, 300 kPa = 133.5°C
x2 =(v2-vf)/(vg-vf) = (0.22127-0.001073)/(0.60582-0.001073) = 0.364
u2 =uf + x2*ufg = 561.11 + (0.364*1982.1) = 1282.8 kJ/kg
-Qout = (2*(1282.8-2793.7)) + (3*(1282.8-1595.4)) = -3959.6 kJ
Qout = 3959.6 kJ
<span>Accretionary
wedge. The accretionary wedge is formed from several sediments: marine
sediments, but also erosional materials from volcanoes. ocean floor basalts,
volcanic island masses, pelagic sediments, trench sediments and other materials
transported by gravity. All these sediments are in the non-subducting tectonic
plate of the Earth.</span>