Answer:
θ = 41.8º
Explanation:
This is an internal total reflection exercise, the equation that describes this process is
sin θ = n₂ / n₁
where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂
θ = sin⁻¹ n₂ / n₁
let's calculate
θ = sin⁻¹ (1.00 / 1.50)
θ = 41.8º
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
matter = solid, liquid gas.
energy typyes, kinetc in a gas, kinetc and electrostatic in solids and liquids