Answer:
c)At a distance greater than r
Explanation:
For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance between the satellite and the Earth's centre
v is the speed of the satellite
Re-arranging the equation, we write
so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.
Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is
c)At a distance greater than r
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
You can't see beyond a blind turn, so a mirror would allow you to see around the corner.
True because friction happens when two things are rubbed against each other and it creates force and sliding something vigorously against something else can create force.
The answer is D. I hope this helps
