Answer:Yes
Explanation:yeah i think it could be possible because the universe is so far and wide its probably infinite and if we did i personally believe that we would be absolutely be annihilated by them because we would be so far behind technology wise we would have no use for the\
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YOU FELL FOR IT FOOL THUNDER CROSS SPLIT ATTACK</h2><h2>
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Answer:
Local atmospheric pressure in kPa = 97.932 kPa
Local atmospheric pressure in mm Hg = 734.78 mm Hg
Explanation:
Given:
= absolute pressure = 46 kPa = 46000 kPa
= density of air = 
= density of mercury = 
= altitude of the plane = 6400 m
Assume:
= pressure decrease due to air at height h
= local atmospheric pressure
As we go up the altitude, the pressure decreases. So, the reading of the pressure will be equal to the decrease in pressure due to height subtracted from the local atmospheric pressure.
Now, conversion of the local atmospheric pressure is quite simple. let us assume that height of H mm of Hg is required to exert a pressure equivalent to the local atmospheric pressure.
Hence, 734.78 mm Hg is the local atmospheric pressure in mm Hg.
Answer:
70 N
21°
1.1 m/s²
Explanation:
Draw a free body diagram of the block. There are three forces:
Weight pulling straight down
Normal force pushing perpendicular to the incline
Friction force pushing parallel to the incline
Part 1
Sum the forces in the perpendicular direction:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
The block is at rest, so F = N μs:
F = N μs
F = mg μs cos θ
F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)
F = 70 N
Part 2
Sum the forces in the parallel direction (down the incline is positive):
∑F = ma
mg sin θ − F = 0
mg sin θ = N μs
mg sin θ = mg μs cos θ
tan θ = μs
θ = atan μs
θ = atan 0.38
θ = 21°
Part 3
Sum the forces in the parallel direction (this time, acceleration is not 0).
∑F = ma
mg sin θ − F = ma
mg sin θ − N μk = ma
mg sin θ − mg μk cos θ = ma
a = g (sin θ − μk cos θ)
a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)
a = 1.1 m/s²
