Anet force of 6.8 n accelerates a 31 kg scooter across a level parking lot. what is the magnitude of the scooter’s acceleration? a) 0.22 m/s2b) 0.69 m/s2c) 3.2 m/s2d) 4.6 m/s2

a)0.22 m/s².

Explanation:

Given that

Net force ,F= 6.8 N

mass ,m = 31 kg

From the second law of Newton's

F = m a   ---------------1

Where

F=Net force ,m=mass

a=Acceleration

Now by putting the values in the equation 1

F = m a

6.8 = 31 x a

Therefore the acceleration of the scooter will be 0.22 m/s².

The answer will be "a".

a)0.22 m/s².

massxgravity x heght=weightxheight= pe

4900=490xh

h=10