For a 650-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth’s mean radius, the resultant values are is mathematically given as
- V=5.59*10^3m/s
- T=3.98hours
- F=1.47*10^3N
<h3>What are the satellite’s orbital speed, the period of its revolution, and the gravitational force acting on it.?</h3>
Generally, the equation for satellite orbital speed is mathematically given as
a)
V=5.59*10^3m/s
b)
T=3.98hours
c)
Therefore
F=1.47*10^3N
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Answer:
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The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of
950 N = (2300 kg) <em>a</em>
<em>a</em> = (950 N) / (2300 kg)
<em>a</em> ≈ 0.413 m/s²
Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of
<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)
<em>v</em> = -0.75 m/s
which means the wind slows the boat down to a velocity of 0.75 m/s westward.