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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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B

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3 years ago
If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged
kirill115 [55]

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      θ’ = θ₀ / 2

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The first minimum occurs for m = 1, so the diffraction equation of a slit remains

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in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

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in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

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In this exercise we are told that the opening changes

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we substitute

          θ ‘= 1.22  λ / 2a

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