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Lorico [155]
3 years ago
15

At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev

el density of air to be 1.29 kg/m3, how fast must it move over the upper surface to create the ideal lift?
Physics
1 answer:
olganol [36]3 years ago
5 0

Answer:

the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

<em>v₂ = 73.4 m/s </em>

<em></em>

Therefore, the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

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Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the
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Answer: 1.8\°

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d=3.4(10)^{-5}m is the width of the slit

\lambda=540 nm=540(10)^{-9}m is the wavelength of the light  

n is an integer different from zero.

Now, the second-order diffraction angle is given when n=2, hence equation (1) becomes:

dsin\theta_{2}=2\lambda (2)

Now we have to find the value of \theta_{2}:

sin\theta_{2}=\frac{2\lambda}{d} (3)

Then:

\theta_{2}=arcsin(\frac{2\lambda}{d})   (4)

\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})   (5)

Finally:

\theta_{2}=1.8\°   (6)

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What are the comparisons and contrast of adaptation and natural selection? ASAP!!!!!!
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3 years ago
A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r
katrin2010 [14]

Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force(F)/Area(A)

While ε_z is longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

6 0
3 years ago
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