Answer: 28.96 V
Explanation:
Given
No of loops on the armature, N = 80
Length of the loop, l = 12 cm = 0.12 m
Width of the loop, b = 8 cm = 0.08 m
Speed of the armature, 1200 rpm
Magnetic field of the loop, B = 0.30 T
To solve this, we use the formula
V(max) = NBAω
Where,
A = area of loop
A = l*b = 0.12 * 0.08
A = 0.0096 m²
ω = 1200 rpm = 1200 * 2π/60 rad/s
ω = (1200 * 2 * 3.142) / 60
ω = 7540.8 / 60
ω = 125.68 rad/s
Substituting the values into the formula
V(max) = NBAω
V(max) = 80 * 0.30 * 0.0096 * 125.68
V(max) = 80 * 0.362
V(max) = 28.96 V
Therefore, the maximum output voltage of the generator would be 28.96 V
You could use grams hope this helps
Answer:
i did not under stood why did you write you'd tell me a lie.(add a suitable tag)
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
He was the first to improve the telescope i believe
