Answer:
A) At point 1, local acceleration = 0.5 m/s²
At point 2, local acceleration = 1.0 m/s²
B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Explanation:
Local acceleration at those points is the instantaneous acceleration at those points and it is given as
a = dv/dt
At point 1, v₁ = 0.5 t
a₁ =dv₁/dt = 0.5 m/s²
At point 2, v₂ = 1.0 t
a₂ = dv₂/dt = 1.0 m/s²
b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time
Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t
Time = t
Average acceleration = 0.5t/t = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
I think it's 'C' but I won't know for sure until you let me see the diagram.
Answer:
2 m/s
Explanation:
From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.
Let the 10.00 kg mass be
and the 12.0 kg mass be
. When they collide and stick, they have a combined mass of
.
Momentum is given by
. Set up the following equation:
, where
is the desired final velocity of the masses.
Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.
Solving yields:

We can’t see the attachment :(