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Marina CMI [18]
3 years ago
6

The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For

example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.
Physics
1 answer:
serg [7]3 years ago
7 0

Answer:

- d = 2; m = 2; m = 2

- d= 1; m = 1; m = 2

- d = 1; m = 1; m = 1

- d = 2; m = 1; m = 2

- d = 2; m = 1; m = 1

Explanation:

The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.

- d = 2; m = 2; m = 2

- d= 1; m = 1; m = 2

- d = 1; m = 1; m = 1

- d = 2; m = 1; m = 2

- d = 2; m = 1; m = 1

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2. A jack exerts a vertical force of 4.5 X 103
skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

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\\  \\

Given :-

\star  \sf  \small force = 4.5 \times  {10}^{3}  \: newton

\star  \sf  \small distance = 0.25 \: meter

\\  \\

To find:-

\sf \star \: work = \: ?

\\  \\

Solution:-

we know :-

\bf \dag \boxed{ \rm work = force \times distance}

\\  \\

So:-

\dashrightarrow \sf work = force \times distance

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{0 \cancel.5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \cancel \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work =  \dfrac{4\cancel.5}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{3 - 1} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{1}  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times  \cancel{10}}{ \cancel2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times 5}{ 1} \\

\\  \\

\dashrightarrow \sf work =225 \times 10

\\  \\

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vladimir2022 [97]

Answer:

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Explanation:

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galben [10]

Answer:

C...................

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On planet X, an object weighs 7.04 N. Onplanet B where the magnitude of the free-fallacceleration is 1.91g(whereg= 9.8 m/s2isthe
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<h2>Mass of object in Earth is 1.37 kg</h2>

Explanation:

On planet B where the magnitude of the free-fall acceleration is 1.91g , the object weighs 25.74 N.

We have

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On planet B

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Mass is constant for an object. It will not change with location.

Mass of object in Earth = Mass of object in Planet B

Mass of object in Earth = 1.37 kg

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4 0
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