Answer
given,
time = 10 s
ship's speed = 5 Km/h
F = m a
a is the acceleration and m is mass.
In the first case
F₁=m x a₁
where a₁ = difference in velocity / time
F₁ is constant acceleration is also a constant.
Δv₁ = 5 x 0.278
Δv₁ = 1.39 m/s
a₁ = 0.139 m/s²
F₂ =m x a₂
F₃ = F₂ + F₁
Δv₃ = 19 x 0.278
Δv₃ = 5.282 m/s
a₃=Δv₂ / t
a₃ = 0.5282 m²/s
m a₃=m a₁ + m a₂
a₃ = a₂ + a₁
0.5282 = a₂ + 0.139
a₂=0.3892 m²/s
F₂ = m x 0.3892...........(1)
F₁ = m x 0.139...............(2)
F₂/F₁
ratio =
ratio = 2.8
Answer:
will be less than
and
will be greater than
.
Explanation:
As we know from the conservation of mass, the rate at which any amount of fluid mass () is entering in a system is equal to the rate at which the same amount of fluid mass (
) is leaving the system.
Rate of mass flow can be written as,
where is the density of the fluid,
is the area through which the fluid is flowing and
is the velocity of the fluid.
Now, according to the problem, as the density of the fluid does not change, we can write
where and
are the cross-sectional areas through which the fluid is passing and
and
are the velocities of the fluid through the respective cross-sectional areas.
As according to the problem, , so from the above formula
.
Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.
So, as in this case the pressure in the large cross-sectional area
will be greater than the pressure
in the small cross sectional area, i.e.,
.