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Kryger [21]
3 years ago
10

When you hear an echo you are hearing a ________of the original sound wave

Physics
1 answer:
krek1111 [17]3 years ago
7 0

Answer:

reflection

Choice D is correct

Explanation:

An echo is simply a reflection of the original sound wave which arrives at the listener with a delay after the original sound.  Sound waves can be reflected, that is bouncing off objects in the same manner a solid object can bounce off the ground. The echo sounds entirely the same as the original sound but the direction of the sound is altered.

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If you kicked your mom <br><br> would she be mad?
vfiekz [6]

Answer:

Yes !

Explanation:

8 0
2 years ago
A monkey (mass m) is swinging on a vine of length L while carrying a bunch of bananas (a large bunch, mass m/2). His swinging mo
dmitriy555 [2]

Answer:

Explanation:

The period of oscillation will remain unchanged because the period of oscillation of a pendulum does not depend upon the mass of the bob  . Here monkey along with bunch of banana represents bob .

When the monkey and banana were at height h /2 , they have potential energy as well as kinetic energy . banana is separated from the system . It carried its total energy along with it . But the energy of monkey remained intact with it . So it will keep on moving as usual . So it will attain the same maximum height as before .

Hence the amplitude of oscillation too will remain unchanged .

6 0
2 years ago
Kinetic energy of an object is equal to
Arte-miy333 [17]
Choice-'b' says the formula for kinetic energy in words.

     KE = (1/2) · (M) · (S²)
7 0
3 years ago
Read 2 more answers
A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\
zloy xaker [14]

Answer:

  t = 1.77 s

Explanation:

The equation of a traveling wave is

       y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

      v = λ f

Where the frequency and period are related

     f = 1 / T

we substitute

      v = λ / T

let's develop the initial equation

    y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

    y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

 

         2π /λ = 1.65

          λ = 2π / 1.65

          λ = 3.81 m

         2π / T = 4.64

          T = 2π / 4.64

          T = 1.35 s

we seek the speed of the wave

           v = 3.81 / 1.35

           v = 2.82 m / s

           

Since this speed is constant, we use the uniformly moving ratios

          v = d / t

           t = d / v

           t = 5 / 2.82

           t = 1.77 s

3 0
3 years ago
Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi
kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors \vec{A} and \vec{B} gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors \vec{A} = \vec{P}-\vec{Q} and \vec{B} = \vec{R} - \vec{Q} are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)

\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)

\vec{A} \times  \vec{B} = (A_y B_z - B_y A_z) \  \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}

\vec{A} \times  \vec{B} = ( (-1) * 3 - 1 * (-3) ) \  \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}

\vec{A} \times  \vec{B} = ( - 3 + 3 ) \  \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}

\vec{A} \times  \vec{B} = 0 \  \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}

\vec{A} \times  \vec{B} =(0, -33, 12)

<h2>B)</h2>

We know that \vec{A} and \vec{B} are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

|\vec{A} \times  \vec{B} | = 2 * area_{triangle}

so:

\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}

\sqrt{1233} = 2 * area_{triangle}

35.114= 2 * area_{triangle}

17.55 \ units \  of \ area =  area_{triangle}

5 0
2 years ago
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