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3 years ago

How does a helicopter fly ? ( I need it explained in the view of physics)

2 answers:

6 0

The science of a helicopter is exactly the same as the science of an airplane: it works by generating lift—an upward-pushing force that overcomes its weight and sweeps it into the air. Planes make lift with airfoils (wings that have a curved cross-section).

Zielflug [23.3K]3 years ago

3 0

Answer:

because blades of helicopter create a high pressure difference between lower and higher portion of blade so due to this pressure difference helicopter became able to fly.

Explanation:

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A small rubber ball is thrown at a heavier, larger basketball that is still. The small ball bounces off the basketball. Assume t

meriva

Part A.
The forces are the same because the force from the smaller ball it transferring its Energy through the basketball and it's rebounding as Connecticut Energy back up to the smaller ball

6 0

3 years ago

Read 2 more answers

Which color of light is diffracted at a greater angle from a diffraction grating, red or yellow light?

topjm [15]

The amount of diffraction depends on the wavelength of light, with shorter wavelengths being diffracted at a greater angle than longer ones (in effect, blue and violet<span> light are diffracted at a larger angle than is red light).

I hope my answer has come to your help. God bless and have a nice day ahead!
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3 0

3 years ago

A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a

kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

$\lambda = 720 \ nm$

or,

$= 720\times 10^{-9} \ m$

$D=0.85 \ m$

$d = 0.22 \ mm$

or,

$=0.22 \times 10^{-3} \ m$

As we know,

Fringe width is:

⇒ $\beta=\frac{\lambda D}{d}$

hence,

Separation between second and third bright fringes will be:

⇒ $\theta=\beta=\frac{\lambda D}{d}$

$=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}$

$=2.78\times 10^{-3} \ m$

or,

$=2.78 \ mm$

8 0

2 years ago

To move a heavy object, like a refrigerator what could be used to help decrease the frictional force?

Olenka [21]

Answer:

A. Remove everything in the refrigerator to lighten the load.

B. Put a lubricant between the surface of the object and the floor

C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily

Explanation:

Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.

F = μN where N is the normal force which depends on the mass.

Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.

8 0

3 years ago

Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine

TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by $tanθ = y/D$

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0

3 years ago