Answer:
Explanation:
The magnitude of the acceleration makes an angle of 30° with the tangential velocity.
Resolving the acceleration to tangential and radial acceleration
at = aCos30 = √3a/2
ar = aSin30 = ½a
a = 2•ar
Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:
at = Rα
Then, α = at/R
since at = √3a/2
Then, α = √3 at/2R, equation 1
The radial acceleration is given as
ar = ω²R
Note that, at² + ar² = a²
at = √(a²-ar²)
Back to equation 1
α = √3 at/2R
α = √3√(a²-ar²)/2R
α = √3√(a²-(w²R)²)/2R
α = √3(a²-w⁴R²) / 2R
Also, a = 2•ar = 2w²R
Then,
α = √3((2w²R)²-w⁴R²) / 2R
α = √3(4w⁴R²-w⁴R²) / 2R
α = √3(3w⁴R²) / 2R
α = √9w⁴R² / 2R
α = 3w²R / 2R
α = 3w²/2
<span>Air resistance will eventually equal the thrust force, so there is no resultant force meaning the car will stay at the same speed and can no longer accelerate. </span>
Answer:
They are in free-fall motion.
Explanation:
The Earth orbiting astronauts are falling at an acceleration that is the same or greater than the acceleration due to gravity i.e., 9.81 m/s². If you are continuously falling at this rate then you will feel weightless.
This same effect is felt while going down in an elevator. When you down in an elevator you feel that you are lighter and feel that something is pushing you up. Earth-orbiting astronauts feel the same effect but the accelration is greater hence they feel weightless.
Answer:
= 3521m/s
The tangential speed is approximately 3500 m/s.
Explanation:
F = m * v² ÷ r
Fg = (G * M * m) ÷ r²
(m v²) / r = (G * M * m) / r²
v² = (G * M) / r
v = √( G * M ÷ r)
G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴
r = 32000km = 32 * 10⁶ meters
G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶
v = √1.24 * 10⁷
v = 3521.36m/s
The tangential speed is approximately 3500 m/s.
