1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
BartSMP [9]
2 years ago
5

What beat frequencies are possible with tuning forks of frequencies 255, 258, and 260 hz ?

Physics
1 answer:
zavuch27 [327]2 years ago
7 0

Possible beat frequencies with tuning forks of frequencies 255, 258, and 260 Hz are 2, 3 and 5 Hz respectively.

The beat frequency refers to the rate at which the volume is heard to be oscillating from high to low volume. For example, if two complete cycles of high and low volumes are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. A common physics demonstration involves producing beats using two tuning forks with very similar frequencies. If a tine on one of two identical tuning forks is wrapped with a rubber band, then that tuning forks frequency will be lowered. If both tuning forks are vibrated together, then they produce sounds with slightly different frequencies. These sounds will interfere to produce detectable beats. The human ear is capable of detecting beats with frequencies of 7 Hz and below.

A piano tuner frequently utilizes the phenomenon of beats to tune a piano string. She will pluck the string and tap a tuning fork at the same time. If the two sound sources - the piano string and the tuning fork - produce detectable beats then their frequencies are not identical. She will then adjust the tension of the piano string and repeat the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency will be reduced and approach 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; that is, they play the same frequency. The process allows a piano tuner to match the strings' frequency to the frequency of a standardized set of tuning forks.

Learn more about  beat frequency here : brainly.com/question/14157895

#SPJ4

You might be interested in
What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a
Elena L [17]

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

6 0
3 years ago
what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s
kotegsom [21]
Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.
7 0
3 years ago
Which line represents a stationary object?
sineoko [7]

Answer:

I think its B but I may be wrong

3 0
3 years ago
Read 2 more answers
Which of the following is an example of deposition?
Murrr4er [49]
 B would be the correct answer 
4 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
Other questions:
  • What is required to cause acceleration
    6·2 answers
  • A wagon has a mass of 100 grams and an
    12·1 answer
  • Which of the following materials is an insulator against electric current
    7·2 answers
  • List and define three types of intermolecular forces and identify which types of molecules each forces affects.
    8·1 answer
  • In what direction would you head from New York state to find warmer temperatures
    5·1 answer
  • A ball is thrown from 1 m above the ground. The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal. What
    14·1 answer
  • Se quiere calcular la viscosidad de un líquido por el método de Oswald y se obtienen los siguientes resultados:
    15·1 answer
  • Review the vocabulary associated with nuclear and wave therapies
    12·1 answer
  • This equation goes with which law?
    10·1 answer
  • URGENT!! ILL GIVE BRAINLIEST! AND 100 POINTS
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!