Answer:
115 kPa
Explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
Assuming no elevation change, h₁ = h₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
Plugging in values:
(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²
P = 115,000 Pa
P = 115 kPa
Answer:
The answer is below
Explanation:
Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.
The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph
vₐ = vₙ + vₐₙ
vₐ = vₐₙ
Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.
The direction of the wind θ is:
sin(θ / 2) = vₙ / 2vₐ
sin(θ / 2) = 70/ (2*135)
sin(θ / 2) = 0.2593
θ / 2 = sin⁻¹(0.2593) = 15
θ = 30⁰
2α = 180° - 30°
2α = 150°
α = 75°
a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.
Answer:
2.85 s .
Explanation:
y(t) = y(0) + v₀t + 1/2 gt²
y(t) is vertical displacement , y(0) is initial position , v₀ is initial velocity and t is time required to make vertical displacement and g is acceleration due to gravity.
Here y(0) is zero , v₀ = 14 m/s , g = 9.8 m s⁻² , y(t ) = 0 , as the pumpkin after time t comes back to its initial position, that is ground .
We shall take v₀ as negative as it is in upward direction and g as positive as it acts in downward direction
Put the values in the equation above,
0 = 0 - 14t + 1/2 x 9.8 t²
14 t = 1/2 x 9.8 t²
t = 28 / 9.8
t = 2.85 s .
