Answer:
70.7m/s
Explanation:
V = ( v + u)
_____
2
V = average velocity
v = final velocity
u = initial velocity
velocity = distance
______
time
u (initial velocity) = 33 ÷ 2
change mins to seconds
1 min = 60 secs
2 mins = ?
cross multiply
60 × 2 = 120 secs
u = 33 ÷ 120
u = 0.275m/s
v ( final velocity) = 29 ÷ 1.3
change mins to secs
1 min = 60 secs
1.3 mins = ?
cross multiply
1.3 × 60 = 78 secs
v = 29 ÷ 78
v = 0.371 m/s
V = ( v + u )
_____
2
V = ( 0.371 + 0.275)
____________
2
V = 0.51 m/s
140 ? i’m guessing there doesn’t look like there’s much to the question
Answer:
= 3,126 m / s
Explanation:
In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.
the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s
Before the crash
p₀ = m v₁₀ + M v₂₀
After the inelastic shock
= m
+ M
p₀ = 
m v₀ + M v₂₀ = m
+ M
We cleared the end of the train
M
= m (v₁₀ - v1f) + M v₂₀
Let's calculate
3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45
= (-3.9 + 8.82) /3.60
= 1.36 m / s
As we can see, this speed is lower than the speed of the car, so the two bodies are joined
set speed must be
m v₁₀ + M v₂₀ = (m + M)
= (m v₁₀ + M v₂₀) / (m + M)
= (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)
= 3,126 m / s