Answer:
1.324 × 10⁷ m
Explanation:
The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.
Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.
We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have
Rω² = GME/R²
R(2π/T)² = GME/R²
R³ = GME(T/2π)²
R = ∛(GME)(T/2π)²
RE + h = ∛(GMET²/4π²)
h = ∛(GMET²/4π²) - RE
substituting the values of the variables, we have
h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m
h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m
h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m
h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m
h = 19.64 × 10⁶ m - 6.4 × 10⁶ m
h = 13.24 × 10⁶ m
h = 1.324 × 10⁷ m
Answer:
The generation of <u>heat</u>.
Explanation:
Initial velicity Vo.
Sin(23) = 24.7 / Vo
Vo = 24.7/Sin(23)
V0 = 63.2 m/sec
Answer:
D = 30.625 m
Explanation:
given,
Speed of the climber = 1.3 m/s
time = 2.5 s
acceleration due to gravity = 9.8 m/s²
initial speed of the kit = 1.3 m/s
velocity of the kit after 2.5 s
using equation of motion
v = u + a t
v = 1.3 + 9.8 x 2.5
v = 25.8 m/s
distance travel by the kit in 2.5 s
v² = u² + 2 g h
25.8² = 1.3² + 2 x 9.8 x h
19.6 h = 663.95
h = 33.875 m
distance travel by the rock climber in 2.5 s
distance = speed of climber x time
h' = 1.3 x 2.5
h' = 3.25 m
Distance between kit and rock climber
D = h - h'
D = 33.875 - 3.25
D = 30.625 m
The kit is 30.625 m below climber.
