The nuclear force that binds the nucleons together in the atom is relatively
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First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used are:
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m
Answer:
6s
Explanation:
Assume it is dropped from rest and the gravitational acceleration is 10
By the equation of motion under constant acceleration:
180 = (0)t+10(t^2)/2
t = 6 or -6 (rejected)
t = 6 s