Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.
<u>Given the following data:</u>
- Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
- Mass of Moon = 7.36 × 10²² kg
- Distance, r = 4.2 × 10⁶ m.
<h3>How to determine the speed of this satellite?</h3>
In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.
This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:
Fc = Fg
mv²/r = GmM/r²
<u>Where:</u>
- m is the mass of the satellite.
Making v the subject of formula, we have;
v = √(GM/r)
Substituting the given parameters into the formula, we have;
v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)
v = √(1,168,838.095)
v = 1,081.13 m/s.
Speed, v = 1.8 × 10³ m/s.
Read more on speed here: brainly.com/question/20162935
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Answer:
2677.3 N
Explanation:
v₀ = initial speed of the hand = 4.75 m/s
v = final speed of the hand = 0 m/s
m = Total mass of hand and forearm = 1.55 kg
t = time interval for hand to come to rest = 2.75 ms = 0.00275 s
F = Force applied on the leg
Using Impulse-change in momentum equation
F t = m (v - v₀)
F (0.00275) = (1.55) (0 - 4.75)
F = - 2677.3 N
magnitude of force = 2677.3 N
The gentleman bug's angular speed is the same as the ladybug's (1 rev/s)
Answer:
William Ferrel created a tide-prediction machine.
Explanation:
- William Ferrel create a machine in late 19th century that was the best combination of mechanical parts and computer coding.
- It was a mechanical analog computer that could predict the ebb of tides and even the height of tides that could be irregular.
- It was widely used for marine networks and navigation. Later on many improvisations and additional features were added on it.
- During the world war times, this tide prediction machine was of great use for military purpose.
