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Licemer1 [7]
3 years ago
13

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

rs. The top of the window is located 43 meters below the surface. Assuming the specific weight of the water in the pool is 9.8 kN/m^3, what is the resultant force (in kN) on the window? Round to the nearest kN.
Physics
1 answer:
ludmilkaskok [199]3 years ago
4 0

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:

                       y_c = (d + \frac{L}{2} ) \\\\y_c = (43 + \frac{8}{2} ) \\\\y_c = 47 m

- 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.

                         F = ( 9.8 ) * ( 47 )* ( 8 )^2\\\\F = 29479KN

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.

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mina [271]

Answer:

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa

(c) Same steps as in part (a)

P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa

ΔP=9.164-9.14

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The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?
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3 years ago
A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, start
Len [333]

Answer:

a) 16.46 seconds.

b) 13.46 seconds

c) 2.67 m/s²

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Initial velocity = u = 0

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v=u+at\\\Rightarrow 22.22=0+1.35t\\\Rightarrow t=\frac{22.22}{1.35}=16.46\ s

Time taken to accelerate to top speed is 16.46 seconds.

Acceleration = a = -1.65 m/s²

Initial velocity = u = 80 km/h= 80\frac{1000}{3600}=22.22\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=22.22-1.65t\\\Rightarrow t=\frac{22.22}{1.65}=13.46\ s

Time taken to stop the train from top speed is 13.46 seconds

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A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at
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A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed? 

<span>Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. </span>

<span>At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. </span>

<span>So total time is 8.5 + 25.5 or 34 sec. </span>

<span>Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.</span>
7 0
4 years ago
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