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:
This is due to impulse
Explanation:
Impulse equal to mΔv and FΔt
You can set these equal as mΔv = FΔt
When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.
The answer to question one is A.
The answer to question two is A.
The answer to question three is D.
Answer:The acceleration due to gravity g is inversely proportional to the square of the radius in the formula g = GM / R^2 where G is the gravitational constant = 6.67 x 10^-11 Nm^2/kg^2, M is the mass of the Earth and R is the radius of the Earth
Explanation:
Answer:
The answer is 0.83 seconds.
Explanation:
The formula of free fall is following:
Where g=9.8 m/s^2 and t=2 seconds, the rock takes:
19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:
2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds
