Given: Altitude of satellite r = 13,300 Km convert to meter
r = 1.33 x 10⁷ m
Universal Gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²
Mass of the earth Me = 5.98 x 10²⁴ Kg
Required: Period of satellite T = ?
Formula: F = ma; Centripetal acceleration ac = V²/r F = GMeMsat/r²
Velocity of satellite V = 2πr/T
equate T from all given equation.
F = ma
GMeMsat/r² = MsatV²/r cancel Msat and insert V = 2πr/T
GMe/r² = (2πr)²/rT² Equate T or period of satellite
T² = 4π²r³/GMe
T² = 39.48(1.33 X 10⁷ m)³/(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)
T² = 9.29 x 10²² m³/3.99 x 10¹⁴ m³/s²
T² = 232,832,080.2 s²
T = 15,258.84 seconds or (it can be said around 4.24 Hr)
Answer:
No
Explanation:
No matter where an object is, its mass will stay the same, but its weight might change depending on gravity.
The angular speed of the merry-go-round is
ω = 0.10 rad/s
The angular moment of inertia of a mass, m, at a radius, r, from the center of the wheel is
I = mr²
Therefore, the angular moment of inertia for the children are
I₁ = (25 kg)*(1.0 m)² = 25 kg-m²
I₂ = (25 kg)*(1.5 m)² = 56.25 kg-m²
The combined angular momentum is
ω(I₁ + I₂) = (0.10 rad/s)*(25 + 56.25 kg-m²)
= 8.125 (kg-m²)/s
Answer: 8.125 (kg-m²)/s
Answer:
1. t = 3.27 seconds
2. y = 147.3 m
Explanation:
Newton's Laws of Motions.
y = v₁t + 1/2 at²
a = (v₂-v₁)/t
where
y = the vertical distance travelled
v₁ = the initial velocity
v₂ = the final velocity
t = the time
a = the acceleration
final velocity is equal to 0.
So, v₂ = 0.
a = (v₂-v₁)/t
a = (0-30)/t
a = -30/t
plugin values into the first equation:
y = v₁t + 1/2 at²
49 = 30t + 1/2 (-30/t)t²
49 = 30t -15t
49 = 15 t
t = 49/15
t = 3.27 seconds
2.
y = v₁t + 1/2 at²
a = -30/3.27
a = 9.2
y = 30(3.27) + 1/2(9.2) 3.27²
y = 147.3 m
It would be: Speed = Distance / Time
