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icang [17]
3 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]3 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]3 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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3 years ago
Question 5
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3 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
PART A:
olga55 [171]

Answer:

4 m/s^{2}

Explanation:

From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.

The Net force = Mass * Acceleration.

Since Net Force = 20 Newton, Mass = 5kg, therefore;

20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;

Acceleration = \frac{20}{5} which gives 4.

Note: RHS means Right Hand Side.

         LHS means Left Hand Side.

 

7 0
3 years ago
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