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erica [24]
3 years ago
14

When you strike two tuning forks simultaneously, you hear three beats per second. The frequency of the first tuning fork is 440

Hz. What is the frequency of the second tuning fork? 1. 446 Hz 2. 437 Hz 3. 443 ㎐ 4. not enough information is given to decide
Physics
2 answers:
tekilochka [14]3 years ago
5 0

Answer:4

Explanation:

Given

Frequency of beats is 3 beats per second

Frequency of first tuning fork is f_1=440\ Hz

Beat frequency is difference in the frequency of tuning forks i.e.

either f_1-f_2 or f_2-f_1 =3

so f_2=3+440=443\ Hz

or

f_2=440-3=437\ Hz

so there is not enough information given to decide.

AlladinOne [14]3 years ago
3 0

Answer:

The frequency of the second tuning fork is 437 Hz.

(2) is correct option.

Explanation:

Given that,

Frequency of first tuning fork = 440 Hz

Frequency of beat = 3 Hz

We need to calculate the frequency of the second tuning fork

Using formula of beat frequency

beat\ frequency=f_{1}-f_{2}

Put the value in to the formula

3=440-f_{2}

f_{2}=440-3

f_{2}=437\ Hz

Hence, The frequency of the second tuning fork is 437 Hz.

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The frequency of the waves is the reciprocal of the period:

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3 years ago
Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude
natka813 [3]

Answer:

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Explanation:

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So vector u will be u = 10.558 i+12.808 j

Unit vector u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j

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