Answer:
A : hot and moist, maritime tropical
B: cold and dry, maritime polar
C: hot and moist , maritime tropical
D: cold and dry, continental polar
E: hot and moist , maritime tropical
F: cold and dry , maritime polar
Explanation:
Cold air is denser than warm air. The more water vapor that is in the air, the less dense the air becomes. That is why cold, dry air is much heavier than warm, humid air.
Maritime polar (mP) air masses are cool, moist, and unstable. Some maritime polar air masses originate as continental polar air masses over Asia and move westward over the Pacific, collecting warmth and moisture from the ocean.
Maritime tropical (mT) air masses are warm, moist, and usually unstable.
Answer:
300 cos 30 = 40 a + 40 * .2 * 10
Total force = mass * acceleration + frictional force
260 = 40 a + 80
a = 180 / 40 = 4.5 m/s^2
Check:
15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)
T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block
260 - 97.5 = 162.5 net force on 25 kg block
162.5 = 4.5 (25) + 25 * 10 * .2
162.5 = 112.5 + 50 = 162.5
4.5 m/s^2 checks out as correct
Answer:
a)
= 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂) 
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂)
2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ - 
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘=
- u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system
<span>Methods of extraction include: extract by electrolysis, extract by reaction with carbon or carbon monoxide, and extracted by various chemical reactions.</span>