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
Given Information:
Mass = m = 3 kg
Speed = v = 6 m/s
Radius = r = 2 m
Required Information:
Magnitude of the acceleration = a = ?
Answer:
Magnitude of the acceleration = 18 m/s²
Explanation:
The acceleration of the block traveling along a circular path with some velocity is given by
a = v²/r
a = 6²/2
a = 36/2
a = 18 m/s²
Therefore, the magnitude of the acceleration of the block is most nearly equal to 18 m/s².
Bonus:
The corresponding force acting on the block can be found using
F = ma (a = v²/r)
F = mv²/r
<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
</span>
Answer:
If force is applied to cause the motion of the body
Explanation:
In the setup given in this diagram, potential energy can be converted into kinetic energy by applying force on the cart or object to move it down the slope.
Potential energy is the energy due to the position of a body. The body has huge potential energy at its current position.
Kinetic energy is the energy due to the motion of a body.
As the body moves down the slope and its velocity increases, it gains massive kinetic energy.
Down the slope, the kinetic energy increases as the potential energy decreases.
At the bottom of the slope, the potential energy becomes zero and the kinetic energy is at its maximum.
