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Brums [2.3K]
2 years ago
10

What is the difference between a wave and a particle?.

Physics
2 answers:
Nimfa-mama [501]2 years ago
6 0

Answer:

See below.

Explanation:

The difference between these two is that a wave is a disturbance of some quantity in space or intermediate while a particle has a definite mass concentrated on a small area.

pogonyaev2 years ago
4 0
Wave is the disturbance of some quantity in space or intermediate while a particle has a definite mass concentrated on a small area. Wave does not have a dimension of size, only its wavelength and velocity can be calculated. Properties of particles: A particle is a localized point in the space
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A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo
aniked [119]

Answer:

a) 46.5º  b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

     θ = 43.5º

This angle measured from the surface is

     θ_r = 90 -43.5

     θ_s = 46.5º

b) In this part the law of refraction must be used

     n₁ sin θ₁ = n₂. Sin θ₂

     sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

     θ_s = 90 - θ

     θ_s = 90- 43.5

     θ_s = 46.5º

     sin θ₂ = 1 / 1.68  sin 46.5

     sin θ₂ = 0.4318

     θ₂ = 25.6º

The angle with respect to the surface is

     θ₂_s = 90 - 25.6

     θ₂_s = 64.4º

measured in the fourth quadrant

3 0
3 years ago
What is force??????<br> ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀​
8090 [49]

Answer:

Force is an external agency that change or tends to change the position of a body

6 0
3 years ago
The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s
just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c)  78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0
3 years ago
Since many wavelengths of light are measured in nanometers, it's useful to know that planck's constant (h) multiplied with the s
MrRa [10]

Answer:

E = h f      and since f = c / λ     we have E = h c / λ

Thus  λ = h c / E

Also, 1 ev = 1.6E-19 C

λ = 6.63E-34 * 3.0E8 / (1.6 * 1.6E-19)

λ = 6.63 * 3.0 / (1.6 * 1.6) * E-7

λ = 7.77E-7 = 777 nanometers

(B) is correct

4 0
2 years ago
On one of the shelves in your physics lab is displayed an antique telescope. A sign underneath the instrument says that the tele
Vesna [10]

Answer:

The focal length fe of the eyepiece is <em>2.86 cm</em>

Explanation:

Since we are given the telescope's magnification and the length of the tube, we can use the expressions

<em>M = f_o/fe (1)</em> and

<em>l = f_o + fe   (2)</em>

where

  • M is the telescope's magnification
  • l is the length of the tube
  • fe is the focal length of the eye-piece

Rearranging equation (2) to make f_o the subject of the formula, we get

<em>f_o = l - fe</em>

Substituting the above equation into equation (1) we get

<em>M = (l - fe)/fe ⇒ fe = l/(M +1)</em>

<em>                      ⇒ fe = 60/(20 + 1)</em>

                     ⇒ <em>fe  = 2.86 cm</em>

4 0
4 years ago
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