A large meteoroid enters the Earth's atmosphere at a speed of 20.0 km/s and is not significantly slowed before entering the ocea
1 answer:
The shock wave from the meteoroid in the lower atmosphere has a Mach angle of 0.948°.
(a) The meteoroid's speed
Air sound wave speed
Speed of the shock wave in Mach
Hence, 0.948° is the Mach angle of the shock wave from the meteoroid in the lower atmosphere.
<h3>What is the speed of the meteoroid?</h3>A meteoroid's speed can be loosely broken down into three categories: slow, medium, and fast.
- Slow meteors move around the sun at a leisurely pace of about 32 kilometers per second (20 miles per second).
- Medium-speed meteors travel around the sun at approximately 50 kilometers per second (30 miles per second),
- while fast meteors zoom past at over 120 kilometers per second (75 miles per second)!
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Answer:
(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons
(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters
Explanation:
The given parameters are;
The mass of particle, A, m₁ = 2 kg
The mass of particle, B, m₂ = 0.3 kg
The mass of particle, C, m₃ = 0.05 kg
The distance between particle 'A' and particle 'B', r₁ = 0.15 m
The distance between particle 'B' and particle 'C', r₂ = 0.05 m
(a) The gravitational force, 'F', is given as follows;
Where;
F = The force between the two masses
G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²
m₁ = The mass of object 1
m₂ = The mass of object 2
If 'C' is placed at 0.05 m from 'B', we have;
F₂₃ = 6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰
The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)
F₁₃ = 6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰
The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)
The force, 'F', acting on object 'C' = F₁₃ - F₂₃
F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N
The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N
(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x
We have;
F₂₃ = F₁₂
By plugging in the values and removing like terms, we get;
(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²
0.1·x² - 0.23·x + 1.3225 = 0.015·x²
0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0
0.085·x² - 0.23·x + 0.13225= 0
x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))
x ≈ 0.829, or x ≈ 1.877
Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'