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prisoha [69]
3 years ago
15

Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the distance separating objects 1 and 2 is chang

ed to one-half the original value (I.E halved), then the new electrostatic force will be _____ units

Physics
2 answers:
qaws [65]3 years ago
5 0

Answer:

288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.

Explanation:

If all other parameters are constant,

Electrostatic Force of attraction ∝ (1/r²)

F = (k/r²) = 72.0

If r₁ = r/2, what happens to F₁

F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units

Lemur [1.5K]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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Answer:

the magnitude of gravitational force is 6 x 10⁻⁸ N.

Explanation:

Given;

mass of the two people, m₁ and m₂ = 90 kg

distance between them, r = 3.0 m

The magnitude of gravitational force exerted by one person on another is calculated as;

F = \frac{Gm_1m_2}{r^2} \\\\

where;

G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²

F = \frac{Gm_1m_2}{r^2} \\\\F = \frac{6.67\times 10^{-11} \times \ 90 \ \times \ 90}{3^2} \\\\F = 6\times 10^{-8} \ N

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3 years ago
An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular
AURORKA [14]

Answer:

she must increase the current by factor of 7

Explanation:

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B= μo I/(2πr)

where,

'μo'  represents permeability of free space i.e 4π*10-7 N/A2

B=magnetic field

I= current

r=radius

->When r= 1cm=> 0.01m

B1 = μo I_{1/(2π x 0.01)

->when r=7cm =>0.07m

B2 = μo I_{2}/(2π x 0.07)

Now equating both of the magnetic fields, we have

B1= B2

μo I_{1/(2π x 0.01)= μo I_{2}/(2π x 0.07)

I_{1/  I_{2}= 0.01/0.07

I_{1/  I_{2}= 1/ 7

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The circumference of a sphere was measured to be
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To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

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The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

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Maximum error:

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dA = \frac{76}{\pi}cm^2

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\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

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dV = \frac{2888}{\pi^2} cm^3

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