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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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A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
4 years ago
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