What are you asking here?
Given: Mass of earth Me = 5.98 x 10²⁴ Kg
Radius of earth r = 6.37 x 10⁶ m
G = 6.67 x 10⁻¹¹ N.m²/Kg²
Required: Smallest possible period T = ?
Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r
but V = 2πr/T
equate T from all equation.
F = ma
GMeMsat/r² = Msat4π²/rT²
GMe = 4π²r³/T²
T² = 4π²r³/GMe
T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)
T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²
T² = 25,563,909.77 s²
T = 5,056.08 seconds or around 1.4 Hour
Answer:
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Answer: the block at the right lands first
Explanation:
Answer:
- 210 rad/s²
Explanation:
n = frequency of rotation = 3400/60 = 170/3 per sec.
angular velocity ω ( 0 ) at time 0 = 2π n = 2π x 170/3
angular velocity at time t = ω(t) = 0
now, ω²( t) = w²(o) + 2α Φ ( α = angular acceleration and Φ = angular displacement) = 2π x 48 rad.
0 = ( 2π x 170/3 )² + 2α x 48 x 2π
α = - (2π x 170 x 170 )/ (3 x 3 x 2 x 48 ) = 210 rad / s²
