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sesenic [268]
3 years ago
7

A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 H

z. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.
Physics
1 answer:
Ganezh [65]3 years ago
6 0

Answer:

<em> 508Hz</em>

Explanation:

A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - this phenomenon is beat production

frequency is the number of oscillation a wave makes in one seconds.

f1-f2=beats

therefore f1=512Hz

f2=?

beats=4Hz

512Hz-f2=4Hz

f2=512-4

f2=508Hz

the original frequency of the violin is 508Hz

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Which of the following is not a way in which humans directly impact algal bloom production? a. industrial pollution b. wastewate
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d. fishing

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A tow truck drags a stalled car along a road. The chain makes an angle of 30???? with the road and the tension in the chain is 1
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Answer: work = 1,305kJ

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angle= 30°

force= 1,500N

distance= 1,000m

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3 0
3 years ago
A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

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m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

6 0
3 years ago
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