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AlladinOne [14]
3 years ago
6

A standard frequency for tuning musical instruments is 440 Hz for the pitch of A, denoted A4. What is the length, in meters, of

the steel rod that produces the pitch A4 as its fundamental longitudinal resonance?
Physics
1 answer:
Sophie [7]3 years ago
4 0

Answer:

Explanation:

frequency of tuning fork = 440 Hz

wave length λ = velocity of sound  / frequency

= 343 / 440 m

= .7795 m

= 77.95 cm

λ = 77.95 cm

Let length of rod having fundamental wavelength of 77.95 cm be L

L = λ  /2

L = 77.95 / 2

= 38.975 cm .

length of rod = 38.975 cm .

= 0. 39 m

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7 0
4 years ago
Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a lea
lesantik [10]

Answer:

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Question:

Why did you lie about being in college?

6 0
3 years ago
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Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the spe
goblinko [34]

Answer:

the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal is 1835.16 .

Explanation:

We know, wavelength is expressed in terms of Kinetic Energy by :

\lambda=\dfrac{h}{\sqrt{2mE}}

Therefore , E=\dfrac{h^2}{2 \lambda^2 m}

It is given that both electron and proton have same wavelength.

Therefore,

E_e=\dfrac{h^2}{2 \lambda^2 m_e}   .... equation 1.

E_p=\dfrac{h^2}{2 \lambda^2 m_p}   .... equation 2.

Now, dividing equation 1 by 2 .

We get ,

\dfrac{E_e}{E_p}=\dfrac{\dfrac{h^2}{2 \lambda^2 m_e}}{\dfrac{h^2}{2 \lambda^2 m_p}}\\\\\\\dfrac{E_e}{E_p}=\dfrac{m_p}{m_e}

Putting value of mass of electron = 9.1\times 10^{-31}\ kg and mass of proton = 1.67\times 10^{-27}\ kg.

We get :

\dfrac{E_e}{E_p}=\dfrac{1.67\times 10^{-27}\ kg}{9.1\times 10^{-31}\ kg}=1835.16

Hence , this is the required solution.

4 0
4 years ago
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What is any push or pull on an object called?
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Answer:

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Explanation:

6 0
3 years ago
A proton is projected in the positive x direction into a region of a uniform electric field →E =(-6.00 × 10⁵) i^ N/C at t=0 . Th
anzhelika [568]

The acceleration of the proton that is  projected in the positive x direction into a region of a uniform electric field is  5.76×〖10〗^13 m/s^2

The product of the field's strength and the charge's strength yields the magnitude of the electric force acting on a charge traveling in a magnetic field region.

The electric force magnitude acting on the charge is expressed in the equation below.

F=|→E|×|q|

F=|-6.00×〖10〗^5 N/C|×||+1.602×〖10〗^(-19) C|

F=9.612×〖10〗^(-14) N

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Where  F is the force and m is the mass;

Inserting the values into the equation,

a=(9.612×〖10〗^(-14) N)/(1.67×〖10〗^(-27) kg)

a=5.76×〖10〗^13 m/s^2

Therefore, the acceleration of proton is 5.76×〖10〗^13 m/s^2 #SPJ4

brainly.com/question/13263306

#SPJ4

7 0
1 year ago
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