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weqwewe [10]
3 years ago
8

Parallel light waves hit the surface of a still lake and reflect in the same direction. Which interaction of light and matter do

es this illustrate? scattering diffuse reflection refraction regular reflection
Physics
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

Regular reflection

Explanation:

- Reflection is the phenomenon that occurs when a light wave hits the interface between two different mediums and it bounces off back into the same medium. The angle of reflection (measured between the reflected ray and the perpendicular to the interface) is equal to the angle of incidence (measured between the incident ray and the perpendicular to the interface).

There are two different types of reflection:

- Regular reflection: this occurs when the interface between the two mediums is smooth (such as in the case of the still lake), so all the parallel light waves (which have same angle of incidence) are reflected exactly with the same angle of reflection (so, they come out all with same direction)

- Diffuse reflection: this occurs when the interface between the two mediums is not smooth, so each light ray is reflected with a different angle because it hits the interface with a different angle of incidence.

Therefore, in the case of the still lake, the correct answer is regular reflection.

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If the weight of a submerged object is equal to the buoyant force, what net force acts upon the object?
Y_Kistochka [10]

Answer:

A

Explanation:

The weight is acting downwards where as the buoyant force acting upwards (opposite) direction with equal amount of force. so the opposite forces cancel out each other (because of the force amount being equal) and no net force is acting on the object.

Hope i have helped you

Thanks.

7 0
3 years ago
Một vật dao động điều hòa có phương trình là x = 4sin(πt + π/3) (cm; s). Lúc t = 0,5s vật có li độ và vận tốc là
nydimaria [60]

Answer:

0,10s

Explanation:

7 0
2 years ago
What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H
Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 := [H^+]=0.01 mol/L

The concentration of hydrogen ion at pH is equal to 6 : [H^+]'=0.000001 mol/L

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

pH=-\log [H^+]

The hydrogen ion concentration at pH is equal to 2 = [H^+]

2=-\log [H^+]\\

[H^+]=10^{-2}M= 0.01 M=0.01 mol/L

The hydrogen ion concentration at pH is equal to 6 = [H^+]

6=-\log [H^+]\\\\

[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L

Concentration of hydrogen ion at pH is equal to 2 =[H^+]=0.01 mol/L

Concentration of hydrogen ion at pH is equal to 6 = [H^+]'=0.000001 mol/L

The difference between hydrogen ion concentration at pH 2 and pH 6 :

= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L

Moles of hydrogen ion in 0.009999 mol/L solution :

=0.009999 mol/L\times 1 L=0.009999 mol

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

8 0
3 years ago
If you take a deep breath of air, you can float more easily in water because you have _____.
Leni [432]

Answer:

the answer is c

6 0
3 years ago
Read 2 more answers
There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
2 years ago
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