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dezoksy [38]
3 years ago
15

Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other ins

trument
Physics
1 answer:
Lerok [7]3 years ago
8 0

Answer:

259 Hz or 269 Hz

Explanation:

Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).

From the question,

Beat = f₂-f₁................ Equation 1

Note: The frequency of the other instrument is either f₁ or f₂.

If the unknown instrument's frequency is f₁,

Then,

f₁ = f₂-beat............ equation 2

Given: f₂ = 264 Hz, Beat = 5 Hz

Substitute into equation 2

f₁ = 264-5

f₁ = 259 Hz.

But if the unknown frequency is f₂,

Then,

f₂ = f₁+Beat................. Equation 3

f₂ = 264+5

f₂ = 269 Hz.

Hence the beat could be 259 Hz or 269 Hz

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Answer:

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Explanation:

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No. Beta particles are electrons.
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A student using a tuning fork of frequency 512 Hz observes that the speed of sound is 340.0 m/s. What is the wavelength of this
Romashka [77]

Answer:

0.066m

Explanation:

Step one:

given

frequency =512 Hz

The speed of sound is 340.0 m/s.

Required

The wavelength

Step two:

the formula for wavelength is

v=f \lambda

\lambda= v/f

substitute the given data

\lambda= 340/514\\\\\lambda= 0.66m

5 0
3 years ago
hree identical resistors are connected in series. When a certain potential difference is applied across the combination, the tot
pav-90 [236]

Answer:

The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

Explanation:

Given;

three identical resistors connected in series

let the first resistor = R₁

let the second resistor = R₂

let the third resistor = R₃

Rt = R₁ + R₂ + R₃

Since the resistors are identical, thus, R₁ = R₂ = R₃ = R

Rt = 3R

Power is given as;

P = IV = V² / R

P = \frac{V^2}{R_t} = \frac{V^2}{3R} \\\\3P = \frac{V^2}{R} ------equation(i)

If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;

\frac{1}{R_t} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3} \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = \frac{1}{R} +\frac{1}{R} + \frac{1}{R} \\\\\frac{1}{R_t} =\frac{3}{R} \\\\R_t = \frac{R}{3} \\\\P = \frac{V^2}{R_t} = \frac{3V^2}{R} \\\\P_{parallel} = \frac{3V^2}{R} ---------equation (ii)\\\\From \ equation \ (i), 3P_{series} = \frac{V^2}{R}, Substitute \ this \ into \ equation \ (ii)\\\\P = 3(\frac{V^2}{R} )\\\\P = 3(3P)\\\\P_{parallel} = 9P_{series}\\\\P_{parallel} = 9(45)\\\\

P_{parallel} = 405 \ W

Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

4 0
3 years ago
Two infinite wires 20 cm apart each carry a current of 3 A into the paper. d I I d/2 d/2 At a distance d 2 below their midpoint,
Rzqust [24]

<u>Answer:</u>

<h3>As electric current is carried in a cable, around it, a magnetic field is created. The lines of the magnetic fields form concentric circles around the wire. The direction of the magnetic field hinges on the direction of the current. It can be calculated by pointing the thumb of your right hand in the direction of the moment, using the "right hand law." The position of your curled fingers is in the magnetic field lines. The magnetic field magnitude depends on the sum of current, and the distance from the wire carrying the charge.</h3>

<u></u>

<u>Explanation:</u>

Determine the direction of vector B magnitude B: B: B=\mu_{0} * 1 /(2 \pi r): r=d / 2 * \sqrt{2}

\cos \alpha=1 / 2 \Rightarrow \alpha=450

Resultant magnitude strength: B=2 B^{*} \cos \alpha=B=u_{0}^{*} 1 /(2 \pi r)=4.24^{*} 10^{-6} T=4.24 u T its direction is pointing to the left.

Note: Refer the image attached below

3 0
3 years ago
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