PE=mgh
180=0.5*10*h
180=5h
h=180/5=36 m
Answer:
1500 m/s
Explanation:
Recall that for a wave,
Speed = frequency x wavelength
here we are given frequency = 500 Hz and wavelength = 3m
simply substitute into above equation
Speed = 500 Hz x 3m
= 1500 m/s
Answer:
a. 13.7 s b. 6913.5 m
Explanation:
a. How much time before being directly overhead should the box be dropped?
Since the box falls under gravity we use the equation
y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.
So,
y = ut - 1/2gt²
y = 0 × t - 1/2gt²
y = 0 - 1/2gt²
y = - 1/2gt²
t² = -2y/g
t = √(-2y/g)
So, t = √(-2 × 919 m/-9.8 m/s²)
t = √(-1838 m/-9.8 m/s²)
t = √(187.551 m²/s²)
t = 13.69 s
t ≅ 13.7 s
So, the box should be dropped 13.69 s before being directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m
Answer:
L=31.9 mm
δ = 0.22 mm
Explanation:
Given that
v= 14 m/s
ρ=997 kg/m³
μ= 0.891 × 10⁻3 kg/m·s
As we know that when Reynolds number grater than 5 x 10⁵ then flow will become turbulent.
L=0.0319 m
L=31.9 mm
The thickness of the boundary layer at that location L given as
δ = 0.00022 m
δ = 0.22 mm
