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Oksana_A [137]
4 years ago
9

Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed

from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere?
Physics
1 answer:
quester [9]4 years ago
4 0

Answer:

F = - 1,598 10⁻³ N

Explanation:

Electic strength is given by Coulomb's law

          F = k q₁ q₂  / r²

Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges

In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign

       q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹

       q₀ = 4.0 10⁻⁶ C

Let's calculate

       F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²

       F = - 1,598 10⁻³ N

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Answer:

The frequency would double.

Explanation:

Given:

Speed of wave (v) = constant.

Frequency of wave initially (f₁) = 2 Hz

Initial wavelength of the wave (λ₁) = 1 m

Final wavelength of the wave (λ₂) = 0.5 m

Final frequency of the wave (f₂) = ?

We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.

Therefore, framing in equation form, we have:

Wavelength × Frequency = Speed

\lambda\times f=v

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\lambda\times f=constant\ or\ \\\lambda_1\times f_1=\lambda_2\times f_2

Now, plug in the given values and solve for 'f₂'. This gives,

1\times 2=0.5\times f_2\\\\f_2=\frac{2}{0.5}=4\ Hz

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Motors 1 Horsepower and under are made in certain step sizes like

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2 years ago
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Answer:

--

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