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viktelen [127]
3 years ago
9

Two objects that may be considered point masses are initially separated by a distance d. The separation distance is then decreas

ed to d/4. How does the gravitational force between these two objects change as a result of the decrease?
Physics
1 answer:
ozzi3 years ago
8 0

Answer:

Increased by 16 times

Explanation:

F = Gravitational force between two bodies

G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/kg s²

m₁ = Mass of one body

m₂ = Mass of other body

d = distance between the two bodies

F=\frac{Gm_1m_2}{d^2}\\ F=\frac{1}{d^2}\quad \text {(as G and masses are constant)}

F_{new}=\frac{1}{\left (\frac{d}{4}\right )^2}\\\Rightarrow F_{new}=\frac{1}{\frac{d^2}{16}}\\\Rightarrow F_{new}={16}\times \frac{1}{d^2}\\\Rightarrow F_{new}=16\times F

∴Force will increase 16 times

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Answer:

The capacitance per unit length is 5.06\times10^{-11}\ F/m

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

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Using formula of charge per length

\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}

Put the value into the formula

\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}

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We know that,

\lambda=\dfrac{Q}{L}

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Using formula of capacitance per unit length

C=\dfrac{\dfrac{Q}{L}}{\Delta V}

C=\dfrac{6.068\times10^{-9}}{120}

C=5.06\times10^{-11}\ F/m

Hence, The capacitance per unit length is 5.06\times10^{-11}\ F/m

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Snezhnost [94]

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