Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s
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.,
.
In order to solve this problem, it is necessary to apply the concepts related to the Pressure according to the Force and the Area as well as to the pressure depending on the density, gravity and height.
In the first instance we know that the pressure can be defined as
Where
F= Force
A = Mass
In the second instance the pressure can also be defined as
Where,
Density of Fluid at this case Water
g = Gravitational Acceleration
h = Height
If we develop the problem to find the pressure then,
Through the second equation we can find the depth to which it can be submerged,
Re-arrange to find h