A geosynchronous satellite orbits above the equator of Earth in a circular orbit, remaining in a fixed position with respect to
1 answer:
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Explanation:
It is given that,
The distance from the pitcher's mound to the batter is 43 feet, d = 43 feet = 0.00814 miles
Speed with which ball leaves the ball, v = 110 mph
Let t is the time elapses between the hit and the ball reaching the pitcher. It is given by :
t = 0.000074 hours
or
t = 0.2664 seconds
So, the time between the hit and the ball reaching the pitcher is 0.2664 seconds. Hence, this is the required solution.
Answer:
At the closest point
Explanation:
We can simply answer this question by applying Kepler's 2nd law of planetary motion.
It states that:
"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"
In this problem, we have a comet orbiting around the Sun:
- Its closest distance from the Sun is 0.6 AU
- Its farthest distance from the Sun is 35 AU
In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: , therefore if r is larger, then v (velocity) must be lower).
On the other hand, when the the comet is closer to the Sun the line must move faster (, if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.