Answer: 5,640 s (94 minutes)
Explanation:
the tangential speed of the HST is given by
(1)
where
is the length of the orbit
r is the radius of the orbit
T is the orbital period
In our problem, we know the tangential speed: 
. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:
So, we can re-arrange equation (1) to find the orbital period:
Dividing by 60, we get that this time corresponds to 94 minutes.
The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
Answer:
The slope of a graph of position vs time
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