Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
"(1) a satellite moving around Earth in a circular <span>orbit" is the only option from the list that describes an object in equilibrium, since velocity and gravity are working together to keep the orbit constant. </span>
D. They are heterotrophs that digest food internally.
<h2>
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>
Explanation:
One period means time taken to complete one revolution.
In case of swings in one period time it travels the same position through two times.
Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.
For 1 period = 2 times
For 3 periods = 3 x For 1 period
For 3 periods = 3 x 2 times
For 3 periods = 6 times
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.
Answer:
1-D(carbon dioxide, water and sunlight)
2-D(parasitism)
3-C(competition)
Explanation:
hope it helps