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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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3 years ago
Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th
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Answer:

The magnitude of the electric field are 2.38\times10^{4}\ N/C and 1.09\times10^{4}\ N/C

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell q_{inn}=3.50\times10^{-8}\ C

Charge on outer shell q_{out}=1.60\times10^{-8}\ C

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

Where, q = charge

k = constant

r = distance

Put the value into the formula

E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}

E=2.38\times10^{4}\ N/C

The total charge enclosed  by a radial distance 20.5 cm

The total charge is

q=q_{inn}+q_{out}

Put the value into the formula

q=3.50\times10^{-8}+1.60\times10^{-8}

q=5.1\times10^{-8}\ C

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}

E=1.09\times10^{4}\ N/C

Hence, The magnitude of the electric field are 2.38\times10^{4}\ N/C and 1.09\times10^{4}\ N/C

7 0
3 years ago
Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17°cw with respect
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The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

To answer the question, we need to know what polarization of light is.

<h3>What is polarization of light?</h3>

This is when the electric field vector of light is oscillating in one plane.

  • Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
  • Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.
<h3>Intensity of light through each polarized filter</h3>

Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.

So, since the light is initially unpolarized,

  • The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.
  • The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
  • The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
  • The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
  • The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
  • The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
  • The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.
<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>

Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.

So, It/I₀ = 1/2cos¹²17°

= 1/2(0.9563)¹²

= 1/2 × 0.5850

= 0.2925

So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

Learn more about intensity of polarized light here:

brainly.com/question/25402491

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2 years ago
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