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3 years ago

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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

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:

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Two 90.0-kilogram people are separated by 3.00 meters. What is the magnitude of the gravitational force that one person exerts o

VARVARA [1.3K]

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

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

Therefore, the magnitude of gravitational force is 6 x 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:

The magnetic field produced by a steady current flowing in a very long straight wire encircles the wire.In order to solve the question, we use this formula,

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

Therefore, she must increase the current by factor of 7

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3 years ago

What do velocity and acceleration have in common?

Allushta [10]

Acceleration is the change in velocity

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3 years ago

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The circumference of a sphere was measured to be

professor190 [17]

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

$\phi = 76cm$

$Error (dr) = 0.5cm$

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,

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

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

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

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

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

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

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

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

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How might a nation be affected if another nation did not follow agreed upon water rules

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The other nation will get mad at the other nation and they could start a war
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