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Aleks04 [339]
3 years ago
14

When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing

is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?
Physics
1 answer:
Sedaia [141]3 years ago
5 0

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × 10^{-3} slug/ft³

so here P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

0 + \frac{2.048 \times 10^{-3} \times 293.33^2}{2}  = P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}  

solve it we get

P2 = 87.9 lb/ft³

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The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
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a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
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finlep [7]

Answer:

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Explanation:

Given that

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Therefore the acceleration of the clay will be a=24.025\ m/s^2.

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