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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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SOVA2 [1]

Answer:

he wavelength is different (greater) than the wavelength of the incident photon

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To complete the sentence we use the wavelength is different (greater) than the wavelength of the incident photon

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Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons
dybincka [34]
<span>The electric force is given by: 
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 d = Distance between centers of mass 
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 answer
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3 years ago
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What scale can I use for two force 0.8 Newtons and 0.9 Newtons? Someone please help
kirill [66]

Answer:

Explanation:

To plot the given forces of 0.8 N and 0.9 N. It is convenient to take

0.1 N as 1 unit in Graph

For example, 1 unit in the graph is  1 cm

then take 0.1 N equivalent to 1 cm

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A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi
Hitman42 [59]

Answer:

the balls reached a height of 4.9985 m

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mass two M = 2.1 kg

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u = 22 m/s

U = { moving downward} = 12 m/s

Now, using the law conservation of linear moment;

mu + MU = v( m + M )

we solve for "v" which is the velocity of the ball s after collision;

v = (mu + MU) / ( m + M )

so we substitute our given values into the equation

v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )

v = ( 83.6 - 25.2 ) / 5.9

v = 58.4 / 5.9

v = 9.898 m/s

Now, we determine required height using the following relation;

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0 - v² = 2gh

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so we substitute

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h = 4.9985 m

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In the first seconds of flight, the Saturn V rocket achieved an altitude of m, and a velocity of m/s. The rocket weighed approxi
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The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

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