A square window with 80 meter sides is located on the vertical side of a large rectangular pool. The depth of the pool is 5 mete
1 answer:
Answer:
See explanation
Explanation:
Solution:-
- We will first set a datum as the free surface of water in the pool.
- There is a square window with side length ( L = 8 m ) on the vertical side of the pool.
- The depth of the pool is given to be 54 m. The top of the window is ( d = 43 m ) from the free surface.
- We are to determine the resultant hydro static force acting on the window.
- Hydro static force ( F ) acting on an object of area ( A ) fully immersed in fluid is given by the following relationship as follows:
F = Pc*A
Where,
Pc: The hydrostaic pressure acting on the centroid of the obect.
- The hydrostatic pressure acting on the centroid of the object immersed in any fluid can be expressed by the following defining relationship:
Pc = γ*yc
Where,
γ: The specific weight of the fluid
yc: The vertical distance from free-surface to the centroid.
- Assuming homogeneous distribution of material used for the window of square shape. The centroid coincides with the geometric center of the window which is as a distance ( yc ) from the free-surface:
- Now we can calculate the resultant hydro static force ( F ) acting on the window. The specific gravity of fluid ( water ) is γ = 9.8KN/m^3.
Note: The values given in the posted question seem unreasonable. I have assumed values in order of convenience. However, the procedure of solving the problem remains exactly the same.
You might be interested in
The free-body diagram of the forces acting on the flag is in the picture in attachment.
We have: the weight, downward, with magnitude
the force of the wind F, acting horizontally, with intensity
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
By dividing the second equation by the first one, we get
From which we find
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
We have: the weight, downward, with magnitude
the force of the wind F, acting horizontally, with intensity
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
By dividing the second equation by the first one, we get
From which we find
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
The rocket travelled a maximum height at 1.0102 km.
Given,
The acceleration of a rocket (a) = 12 m/s²
The altitude of the rocket (s) = 0.50 km = 0.5×10³m
The maximum height of the rocket (h) = ?
Solution,
A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.
The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²
(a) = Δv/Δt
Where , Δv is change in velocity and Δt is change in time.
The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.
(v)= Δx/Δt
Where,Δx is the change in position and Δt is change in time & v is velocity.
Therefore we know the equation of motion is written as,
v² = u² +2as
Where, v is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.
Then putting the value ,
v² = 0 + ( 2× 10 × 0.5×10³)m/s
v² = m/s
v = 100 m/s
Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,
v² = u² - 2(g)h
h = (v²- u² ) / 2g
h = 10,000/2×9.8
h = 510.2 m
So that the rocket travelled the maximum height ,
(h)= (0.5 km + 510.2m)
(h) = 1.0102 km
Hence, the rocket travelled at the maximum height h is 1.0102 km
To know more about acceleration
#SPJ4