Answer:
Explanation:
a = Δv/Δt = (v₁ - v₀) / (t₁ - t₀)
A a = (10 - 0) / (5 - 0) = 2 m/s²
B a = (10 - 10) / (15 - 5) = 0 m/s²
C a = (5 - 10) / (25 - 15) = - 0.5 m/s²
Both waves would increase right? That seems correct since the water and air temp both equally changed.
given info is... Acceleration(a)=2.6m/s^2
final velocity(v)=26.8m/s
initial velocity(u)=24.6m/s
need to find.... time(t)=?




Explanation:
A=(v2-v1)/t
v2=0,6m/s
v1=2,4m/s
t=4s
a=(0,6-2,4)/4=-1,8/4=-0,45 m/s^2