A jet plane passes over you at a height of 5870 m and a speed of Mach 1.47. (a) Find the Mach cone angle. (b) How long after the
1 answer:
Answer:
(a) Angle is 42.865°
(b) t=12.54s
Explanation:
The jet plane is at 5870m from Earth surface
The jet traveling at speed of 1.47 Mach
Speed of Sound v=343 m/s
Part (a)
As we know :
Substitute the given values
Angle is 42.865°
Part (b)
As the distance given as:
By using some trigonometry
Where h is the height of jet plane
Thus
Substitute the given values
So
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Answer:
a) v = 18.86 m / s, b) h = 8.85 m
Explanation:
a) For this exercise we can use the conservation of energy relations.
Starting point. Like the compressed spring
Em₀ = K_e + U = ½ k x² + m g x
the zero of the datum is placed at the point of the uncompressed spring
Final point. With the spring if compress
Em_f = K = ½ m v²
how energy is conserved
Em₀ = Em_f
½ k x² + m g x = ½ m v²
v² = x² + 2gx
let's reduce the magnitudes to the SI system
m = 500 g = 0.500 kg
x = -45 cm = -0.45 m
the negative sign is because the distance in below zero of the reference frame
let's calculate
v² = 0.45² + 2 9.8 (- 0.45)
v = √355.68
v = 18.86 m / s
b) For this part we use the conservation of energy with the same initial point and as an end point at the point where the rock stops
Em_f = U = m g h
Em₀ = Em_f
½ k x²2 + m g x = m g h
h = ½ x² + x
let's calculate
h = - 0.45
h = 8.85 m
measured from the point where the spring is uncompressed