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kati45 [8]
2 years ago
8

A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 n of lift per square meter of wing.

(the fact that a wing has a top and bottom surface does not double its area.)
Physics
1 answer:
muminat2 years ago
7 0

Answer:

Correct. The lift is directly related to the area of the wing.

Explanation:

The bigger the area of the wing, the bigger the lift will be. But there is a catch: the drag will also be bigger. The denser the air is, more lift and more drag you'll find. At low altitudes, air has more pressure. The higher the pressure is, more lift and drag you'll find. Cold air has more pressure than hot air. Therefore, it's easier for an airplane to takeoff at low altitudes and cold air. But given that drag is also bigger at low altitudes, airplanes tend to have their flight cruises at higher altitudes where there is less lift but there is also less drag as less drag is desirable for saving fuel and flying faster.

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During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 * t2 to so
Ulleksa [173]

8/9 score.....................

3 0
2 years ago
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For a certain optical medium the speed of light varies from a low value of 1.90 × 10 8 m/s for violet light to a high value of 2
Dmitry_Shevchenko [17]

Answer:

a. The refractive index ranges from 1.5 - 1.56

b. 18.7° for violet light and 19.5° for red light.

c. 33.7° for violet light and 35.3° for red light.

Explanation:

a. The refractive index of an object is the ratio of the speed of light in a vacuum and the speed of light in the object.

Mathematically,

n = \frac{c}{v}

The speed of violet light in the object is 1.9 * 10^8 m/s.

The speed of red light in the object is 2 * 10^8 m/s

Hence, the refractive index for violet light is:

n = \frac{3 * 10^8 }{1.9 * 10^8} \\\\n = 1.56

and for red light, it is:

n = \frac{3 * 10^8 }{2 * 10^8} \\\\n = 1.5

Hence, the refractive index ranges from 1.5 - 1.56.

b. The refractive index is also the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

n = \frac{sin(i)}{sin(r)}

The angle of incidence is 30°.

The angle of refraction for violet light will be:

1.56 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.56}  = \frac{0.5}{1.56} \\\\sin(r) = 0.3205\\\\r = 18.7^o

And the angle of refraction for red light will be:

1.5 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.5}  = \frac{0.5}{1.5} \\\\sin(r) = 0.3333\\\\r = 19.5^o

The angle of refraction for red light is larger than that of violet light when the angle of incidence is 30°.

c. The angle of incidence is 60°.

The angle of refraction for violet light will be:

1.56 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.56}  = \frac{0.8660}{1.56} \\\\sin(r) = 0.5551\\\\r = 33.7^o

And the angle of refraction for red light will be:

1.5 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.5}  = \frac{0.8660}{1.5} \\\\sin(r) = 0.5773\\\\r = 35.3^o

The angle of refraction for red light is still larger than that of violet light when the angle of incidence is 60°.

6 0
3 years ago
What is the threshold frequency ν0 of cesium? note that 1 ev (electron volt)=1.60×10−19 j. express your answer numerically in he
andreyandreev [35.5K]

The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

\large {\boxed {E = h \times f}}

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}

\large {\boxed {E = qV + \Phi}}

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem!

<u>Given:</u>

Φ = 2.1 eV = 2 × 1.60 × 10⁻¹⁹ Joule = 3.20 × 10⁻¹⁹ Joule

<u>Unknown:</u>

fo = ?

<u>Solution:</u>

\Phi = h \times fo

3.20 \times 10^{-19} = 6.63 \times 10^{-34} \times fo

fo = ( 3.20 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )

\large {\boxed{fo = 4.83 \times 10^{14} ~ Hz} }

<h3>Learn more</h3>
  • Photoelectric Effect : brainly.com/question/1408276
  • Statements about the Photoelectric Effect : brainly.com/question/9260704
  • Rutherford model and Photoelecric Effect : brainly.com/question/1458544
  • Photoelectric Threshold Wavelength : brainly.com/question/10015690

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

8 0
2 years ago
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Which resultant force is NOT possible if 50 N force and a 60 N force act concurrently?
Sergio [31]

Answer:

The options are not shown, so i will answer in a general way.

Suppose the case where the forces act in opposite directions, then we need to subtract the forces, and we know that the magnitude of the resultant force will be:

60N - 50N = 10N

Now, suppose the case where both forces act in the exact same direction, in that case, we will add the forces to get:

60N + 50N = 110N

Then the only range of forces that we can get in this system, are the forces such:

10N ≤ F ≤ 110N

Any resultant force outside that range is not possible in this situation.

3 0
2 years ago
The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?
aleksandrvk [35]
M = 150 kg.

Final velocity, v = 14 m/s

Initial Velocity, u = 6 m/s

Impulse =  m(v - u)
 
             = 150*(14 - 6)
           
             =  150*8 = 1200 kgm/s  or 1200 Ns 
8 0
3 years ago
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