Answer:
1010713.18851 Pa
387999.259089 N
Explanation:
= Atmospheric pressure =
r = Radius = 2 m
h = Depth = 10 m
= Density of liquid methane =
g = Acceleration due to gravity = 7.44 m/s²
A = Area
Force is given by
The force exerted is 1010713.18851 Pa
The weight of the column of methane is 387999.259089 N
The pressure at a depth is given by
The pressure at the depth is
Cardio respiratory endurance, muscular strength, muscular endurance
Answer:
The magnitude of the lift force L = 92.12 kN
The required angle is ≅ 16.35°
Explanation:
From the given information:
mass of the airplane = 9010 kg
radius of the airplane R = 9.77 mi
period T = 0.129 hours = (0.129 × 3600) secs
= 464.4 secs
The angular speed can be determined by using the expression:
ω = 2π / T
ω = 2 π/ 464.4
ω = 0.01353 rad/sec
The direction
θ = 16.35°
The magnitude of the lift force L = mg ÷ Cos(θ)
L = (9010 × 9.81) ÷ Cos(16.35)
L = 88388.1 ÷ 0.9596
L = 92109.32 N
L = 92.12 kN
Answer:
D) Joe need to walk 481 m in a direction 40.9° north of east
Explanation:
Given:
1. Walk 300 m north
2. Walk 400 m northwest
3. Walk 700 m east-southeast and the treasure is buried there
Taking north as positive y axis and east as positive x axis.
Resolving their positions into x and y vector components.
1. d1 = 300j
2. d2 = -400cos45i + 400sin45j
3 d3 = 700cos22.5i - 700sin22.5j
Resultant position.
d = d1+d2+d3
d = (-400cos45 + 700cos22.5)i + (300+400sin45-700sin22.5)j
d = (363.873)i + (314.964)j
D = √( (363.873)^2 + (314.964)^2)
D =481.25m ~= 481m
Angle = taninverse (314.964/363.873)
Angle = 40.9°
Since the x and y components are both positive, it implies that their position is north of east
Therefore, Joe need to walk 481 m in a direction 40.9° north of east