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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
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
4 years ago
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Explain one way the water cycle affects climate. Use complete sentences.
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Which choice best describes how scientists learn new information during investigations?
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C

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