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icang [17]
2 years ago
10

In the lab activity, you will examine sound waves as they are emitted from a moving source. Predict what will happen to the soun

d waves when the sound source is in motion. Record your prediction (hypothesis) as an “if then” statement. (For example: If you select the GO button, then the train will move)
Physics
2 answers:
irina1246 [14]2 years ago
6 0
<span>As a sound source gets closer, both the volume and the pitch of the sound increased. Then, as the sound source passed by you, both the volume and the pitch of the sound decreased.
 Hope this helps</span>
Serjik [45]2 years ago
5 0

Answer:

Explanation:

If the source is the same, the characteristic sound waves when they are emitted do not change (wave speed), but if the source is in relative motion with the observer, a change in the wavelength must be seen by this relative movement, such as the speed is the same this will be reflected in a change in the frequency heard by the observer

This effect is called the Doppler effect for a moving source is described by the expression, and

     f ’= [(v / (v± vs)] f

With v the speed of sound, vs the speed of the source, f the frequency emitted and f ’the frequency heard, where the sign corresponds to the source away (-) or the source approaching (-)

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Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
What do we mean by the event horizon of a black hole?
TEA [102]
<h2>Answer:</h2>

The event horizon is the surface of a black hole, it is the border of space-time in which the events on one side of it can not affect an observer on the other side.  

That is, at this border also called "point of no return", nothing can escape (not even light) and no event that occurs within it can be seen from outside.

 

5 0
3 years ago
Read 2 more answers
A ball is thrown horizontally from the top of a tall cliff. Neglecting air drag, what vertical distance will the ball have falle
adell [148]

The relevant equation to use here is:

y = v0 t + 0.5 g t^2

where y is the vertical distance, v0 is initial velocity = 0, t is time, g = 9.8 m/s^2

 

y = 0 + 0.5 * 9.8 * 3^2

<span>y = 44.1 meters</span>

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3 years ago
What is the distance an object travels and the direction of its motion from starting point?
castortr0y [4]

Answer: Well you didn't give any answers so my guess would be it depends how much force you put into it and where you throw it.

Explanation:

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