physics question #4?
When you drop a 0.42 kg apple, Earth exerts
a force on it that accelerates it at 9.8 m/s
toward the earth’s surface. According to Newton’s third law, the apple must exert an equal
but opposite force on Earth.
If the mass of the earth 5.98 × 1024 kg, what
is the magnitude of the earth’s acceleration
toward the apple?
Answer in units of m/s
- oubaasLv 77 months ago
accel. a = F/me = 0.42*9.806*10^-24/5.98 = 6.89*10^-25 m/sec^2