Answer:B
Explanation: Gravity pulls denser air and water downward, forcing less dense air and water to move upward. The warm water near the surface of the ocean heats up with sunlight and evaporates, keeping the water cycle in motion.
Answer:
the bearing of the angle AC will be equal to 154°51' 48"
Explanation:
given,
bearing of line AB = 234° 51' 48"
C is anti clockwise angle from 80° from AB
bearing of line AC = ?
To calculate the bearing of line AB 80° anticlockwise movement
bearing of the AC = 234° 51' 48" - 80°
= 154°51' 48"
The bearing can be represented as 154.5148
hence, the bearing of the angle AC will be equal to 154°51' 48"
Answer:
38 m/s
43 m/s
Explanation:
x = 18t + 5.0t²
The instantaneous velocity is the first derivative:
v = 18 + 10.t
At t = 2.0:
v = 18 + 10.(2.0)
v = 38 m/s
The average velocity is the change in position over change in time.
v = Δx / Δt
v = [ (18t₂ + 5.0t₂²) − (18t₁ + 5.0t₁²) ] / (t₂ − t₁)
Between t = 2.0 and t = 3.0:
v = [ (18(3.0) + 5.0(3.0)²) − (18(2.0) + 5.0(2.0)²) ] / (3.0 − 2.0)
v = [ (54 + 45) − (36 + 20.) ] / 1.0
v = 99 − 56
v = 43 m/s
I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
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