Psychoanalytic dream interpretation is a subdivision of dream interpretation as well as a subdivision of psychoanalysis pioneered by Sigmund Freud in the early twentieth century. Psychoanalytic dream interpretation is the process of explaining the meaning of the way the unconscious thoughts and emotions are processed in the mind during sleep.
There have been multiple methods used in psychoanalytic dream interpretation, including Freud's method of dream interpretation, the symbolic method, and the decoding method. The Freudian method is the most prominently used in psychoanalysis and has been for the last century. Psychoanalytic dream interpretation is used mainly for therapeutic purposes in a variety of settings. Although these theories are used, none have been solidly proven and much has been left open to debate among researchers. Some studies have shown that areas of dream interpretation can be invalid and therefore a decline in importance has been seen in psychoanalytic dream interpretation.
When the truck accelerates forward the ball will shift to the back of the bed of the truck because of inertia
Answer:
0.558 atm
Explanation:
We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT
Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:
P₀ = P₁ + P₂ + ....
P₀= total pressure
P₁=P₂= is the partial pressure of each gass
If we can consider that each gas is an ideal gas, then:
P₀= (nRT/V)₁ + (nRT/V)₂ +..
Considering the molecular mass of O₂:
M O₂= 32 g/mol
And also:
R= ideal gas constant= 0.082 Lt*atm/K*mol
T= 65°C=338 K
4.98 g O₂ = 0.156 moles O₂
V= 7.75 Lt
Then:
P°O₂=partial pressure of oxygen gas= (0.156x0.082x338)/7.75
P°O₂= 0.558 atm
Answer: A
Explanation:Earthquakes occur on faults - strike-slip earthquakes occur on strike-slip faults, normal earthquakes occur on normal faults, and thrust earthquakes occur on thrust or reverse faults. When an earthquake occurs on one of these faults, the rock on one side of the fault slips with respect to the other.
Answer:
The second ball lands 1.5 s after the first ball.
Explanation:
Given;
initial velocity of the ball, u = 12 m/s
height of fall, h = 35 m
initial velocity of the second, v = 12 m/s
Time taken for the first ball to land;
determine the maximum height reached by the second ball;
v² = u² -2gh
at maximum height, the final velocity, v = 0
0 = 12² - (2 x 9.8)h
19.6h = 144
h = 144 / 19.6
h = 7.35 m
time to reach this height;
Total height above the ground to be traveled by the second ball is given as;
= 7.35 m + 35m
= 42.35 m
Time taken for the second ball to fall from this height;
total time spent in air by the second ball;
T = t₁ + t₂
T = 1.23 s + 2.94 s
T = 4.17 s
Time taken for the second ball to land after the first ball is given by;
t = 4.17 s - 2.67 s
T = 1.5 s
Therefore, the second ball lands 1.5 s after the first ball.
