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:
33,333.33 N
Explanation:
Given that :
Initial kinetic energy = 500,000 J
Final kinetic energy = 100,000 J
Using the relation :
Force * time = change in momentum (Newton's law)
Force (F) * 0.12 = (500,000 - 100,000)
0.12F = 400,000 J
Force = (400,000 J) / 0.12s
Force = 33333.333
Force = 33,333.33 N
It must be either speeding up, or slowing down, or turning. There are no other possibilities.
PART A)
Equivalent resistance in parallel is given as
now we have
PART B)
since potential difference across all resistance will remain same as all are in parallel
so here we can use ohm's law
As we know i = 7 A current flows through 15 ohm resistance
PART C)
Similarly ohm's law for 20 ohm resistance we can say
He was the first to improve the telescope i believe
