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3 years ago

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A charge of 8.473 nC is uniformly distributed along the x-axis from −1 m to 1 m . What is the electric potential (relative to ze

ro at infinity) of the point at 5 m on the x-axis? The value of the Coulomb constant is 8.98755 × 10^9 N · m2 /C^2 . Answer in units of V.

1 answer:

quester [9]3 years ago

8 0

Answer:

The electric potential is 15.23 V

Explanation:

Given;

charge Q = 8.473 nC = 8.473 x 10⁻⁹ C

position of the charge, r = 5 m

Coulomb constant is 8.98755 × 10⁹ N·m² /C²

Electric potential is given as;

$V= \frac{KQ}{r} = \frac{8.98755 X10^9 X 8.473X10^{-9}}{5}\\\\V = 15.23 V$

Therefore, a charge of 8.473 nC uniformly distributed along the x-axis from −1 m to 1 m, the electric potential (relative to zero at infinity) of the point at 5 m on the same axis is 15.23 V.

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Answer:

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Explanation:

Given that,

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$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

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$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

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Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

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4 years ago

The total mass of the train and its passengers is 750000kg. The train is traveling at a speed of 84m/s. The driver applies the b

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Answer:

|F| = 393750 N

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Given that,

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3 years ago

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Answer:

The two value of the wavelength for the out of tune guitar is

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From the question we are told that

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The difference in beat frequency is $\Delta f = 17.0 \ Hz$

Generally the frequency of the note played by the guitar that is in tune is

$f_1 = \frac{v_s}{\lambda}$

Where $v_s$ is the speed of sound with a constant value $v_s = 343 \ m/s$

$f_1 = \frac{343}{0.0065}$

$f_1 = 5276.9 \ Hz$

The difference in beat is mathematically represented as

$\Delta f = |f_1 - f_2|$

Where $f_2$ is the frequency of the sound from the out of tune guitar

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Answer:

Explanation:

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