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Flauer [41]
3 years ago
6

Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 miles per hour) when the jet stream

descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of 225 m2? Typical air density in Boulder is 1.14 kg/m3, and the corresponding atmospheric pressure is 8.89 ✕ 104 N/m2. (Bernoulli's principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)
Physics
1 answer:
siniylev [52]3 years ago
5 0

Answer:

F = 260 kN

Explanation:

given,                                

speed = 45 m/s            

Area = 225 m²                    

air density = 1.14 Kg/m³                

Atmospheric pressure =  8.89 ✕ 10⁴ N/m²

Using  Bernoulli equation          

assuming the level is same        

P_1 - P_2 = \dfrac{1}{2}\rho V^2

\Delta P= \dfrac{1}{2}\rho V^2          

\Delta P= \dfrac{1}{2}\times 1.14 \times 45^2

\Delta P=1154.25\ Pa

Force = Pressure x area              

Force =1154.25 x 225        

F = 259706.25 N                              

F = 260 kN                      

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<h3>What is induced emf?</h3>

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When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.

The given data in the problem is;

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The maximum emf is found as;

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1 year ago
The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
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Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

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Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

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The magnitude of acceleration of train is calculated as follows.

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              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

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