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Pachacha [2.7K]
3 years ago
13

Object A attracts object B with a gravitational force of 10 Newton’s from a given distance. If the distance between the two obje

cts is doubled, what is the new force of an attraction between them?
Physics
1 answer:
madam [21]3 years ago
7 0

If the distance between two objects is doubled, then the gravitational forces between them become (1/2)² = 1/4 of the original gravitational forces.

If the magnitude of those forces was 10 Newtons before the move then they're 2.5 Newtons after it.

I changed the wording of the discussion to plural, because the wording in the question is misleading.  It shouldn't say "Object A attracts Object B" and then let it go at that.  Gravitational forces ALWAYS occur in equal pairs.  If Object B feels a pull of 10 Newtons toward Object A, then Object A feels a pull of 10 Newtons toward Object B.  

The masses don't even matter.  Gravity is always there equally in both directions.  Your weight on Earth is equal to the Earth's weight on you !

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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

work is done by the jack?

\\  \\

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

\\  \\

\dashrightarrow \bf work =\red{2250\: joule}

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Read 2 more answers
A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N
ANTONII [103]

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

5 0
3 years ago
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It asks for the “Total Mass” basically asking to add, If you add 20 to 125, you get 145! Correct me if im wrong
4 0
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