A=(vf-v₀)/t
Data:
vf=100 m/s
v₀=70 m/s
t=1.5 s
a=(100 m/s - 70 m/s)/1.5 s=30 m/s / 1.5 s=20 m/s².
answer: C. 20 m/s²
Answer:
KE= 1/2mv^2= 1/2*0.0042kg*993m/s= 2.0853joule
Answer:
No.
Explanation:
The force that two particle experience is inversely proportional to the sqare of the distance, this is:
for a distance D
If we move them so that D is doubled:
= 
Then the force they experience is one fourth of the original.
Given
initial position = Xi= 19.9m
Final position Xf = 5.4m
Average velocity= Va = -0.418m/s
it shows displacement is reverse.
To find t=?
As Va = (Xf- Xi) / t
t = (Xf-Xi) / ( Va)
t = ( 5.4-19.9) / (-0.418)
t = (-14.5 ) / (-0.418) (-ve sign cancel out at numerator and denominator)
t =34.69 s