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Vadim26 [7]
3 years ago
15

11. Which of the following phenomena is taking place when sound waves are reflected from a surface along parallel lines? A. Refr

acting B. Absorption C. Focusing D. Diffusion
Physics
1 answer:
allsm [11]3 years ago
4 0
The Answer is C. Focusing             
You might be interested in
Can a body possess velocity at the same time in horizontal and vertical directions?​
iogann1982 [59]

Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

5 0
3 years ago
If the child pulls with a force of 20 N for 12.0 m , and the handle of the wagon is inclined at an angle of 25 ∘ above the horiz
lilavasa [31]

Answer: 217.52 N

Explanation: The applied force is 20 N, the distance covered is 12.0 m and the angle is 25° above the horizontal.

Hence the formulae that defines work done is given by

W = Force × distance

But since the force has been inclined at an angle θ above the horizontal, the horizontal component of force is neccesary to produce the required motion to make the child do work on the wagon.

Hence

Work done = (horizontal component of force) × distance

Work done = F cos θ × distance

Work done = 20 cos 25 × 12 = 217.52 N

4 0
3 years ago
Two conductors made of the same material are connected across the same potential difference. Conductor A has three times the dia
Vitek1552 [10]

Answer:

<em>Good Luck!</em>

<em />

Explanation:

5 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0
3 years ago
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
3 years ago
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