The amount of force applied by individual man to balance the force at the right is 73.2 N.
<h3>Amount of force applied by individual man</h3>
The amount of force applied by individual man to balance the force at the right is calculated as follows;
Thus, the amount of force applied by individual man to balance the force at the right is 73.2 N.
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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:
2.66 m/s² .
Explanation:
Initial velocity , u = 0 m/s
Final Velocity , v = 8 m/s
Time Taken , t = 3 s
So , Acceleration = (v-u)/t = (8 m/s - 0 m/s) /3 sec . = 8/3 m/s² = 2.66 m/s²
distance d = 5 km = 5 x 1000 m = 5000 m
time taken = 25 minute = 25 x 60 sec = 1500 sec
average velocity V = d/t
V = 5000/1500
V = 3.33 m/s towards east
