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3 years ago

When light reaches a barrier or the edge of an opening it diffracts, which means that the light _________.

2 answers:

katrin2010 [14]3 years ago

7 0

The light slightly "bends" as it goes around the edges of the object. This bending is typically dependent upon the size of the barrier in comparison to the wavelength of the object being diffracted. The more similar in size the two are, the much more noticeable the diffraction tends to be.

Reptile [31]3 years ago

3 0

Diffraction is the slight bending of light when it passes around the edge of an object. The amount of bending depends on the relative of the wavelength of the light wave to the size of the opening. If the opening is much larger than the wavelength the bending will be almost unnoticeable but if the two are equal or closer in length then the diffraction will be noted. Therefore, the correct answer is bends.

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Which of the following is most likely to be an observation made by a physiologist

Naya [18.7K]

Do you have the answer choices ?

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3 years ago

Answer and I will give you brainiliest Please heeeelp

Lostsunrise [7]

Answer:

T = 4.905[N]

Explanation:

In order to solve this problem we must perform a sum of forces on the vertical axis.

∑Fy = 0

We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.

$T-m*g=0\\T=0.5*9.81\\T=4.905[N]$

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3 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

3 years ago

Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y

makkiz [27]

Answer:

<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:

1. Light waves are electromagnetic waves, while sound and water waves are mechanical waves, this is the first and principal difference.

2. Electromagnetic waves can propagate in vacuum (they do not need a medium or material), but mechanical waves obligatory need a material to propagate

3. Light waves are always transversal waves, this means the oscillatory movement is in a direction that is perpendicular to the propagation; but mechanical waves may be both: longitudinal waves (the oscillation occurs in the same direction as the propagation) or transversal waves.

4. Electromagnetic waves propagates at a constant velocity (Light velocity) while the velocity of mechanical waves will depend on the type of wave and the density of the medium or material.

5. Mechanical waves are characterized by the regular variation of a single magnitude, while electromagnetic waves are characterized by the variation of two magnitudes: the electric field and the magnetic field

6. Water waves are 2-dimensional waves, while the light and the sound are tridimensional spherical waves

7. Light waves transports energy in the form of radiation, while mechanical waves transport energy with material

3 0

3 years ago

Which of the following is equal to impulse? A. Ft B. m/s C. pt D. mv

tankabanditka [31]

Answer: A (Ft)

Explanation: The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v

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3 years ago