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geniusboy [140]
3 years ago
10

MARKING BRAINLIST | Which situation below would have the STRONGEST gravitational force between them?

Physics
1 answer:
maks197457 [2]3 years ago
6 0

Case d) has the strongest gravitational force

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

a) For this pair of objects:

m1 = 10 kg

m2 = 2 kg

r = 30 km = 30,000 m

So the gravitational force is

F=(6.67\cdot 10^{-11})\frac{(10)(2)}{30000^2}=1.48\cdot 10^{-18}N

b) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 30 km = 30,000 m

So the gravitational force is

F=(6.67\cdot 10^{-11})\frac{(10)(10)}{30000^2}=7.41\cdot 10^{-18}N

c) For this pair of objects:

m1 = 2 kg

m2 = 2 kg

r = 10 km = 10,000 m

So the gravitational force is

F=(6.67\cdot 10^{-11})\frac{(2)(2)}{10000^2}=1.33\cdot 10^{-17}N

d) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 10 km = 10,000 m

So the gravitational force is

F=(6.67\cdot 10^{-11})\frac{(10)(10)}{10000^2}=6.67\cdot 10^{-17}N

Therefore, the  strongest gravitational force is in case d).

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Block 1 (the rightmost block) has

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∑ <em>F</em> = <em>F</em> - <em>T₁</em> - <em>f₁</em> = <em>m₁a</em>

• net vertical force

∑ <em>F</em> = <em>N₁</em> - <em>m₁g</em> = 0

where <em>F</em> = 57.3 N, <em>T₁</em> is the tension in the string connecting blocks 1 and 2, <em>f₁</em> is the magnitude of kinetic friction felt by block 1, <em>m₁</em> = 0.8 kg is its mass, <em>a</em> is the acceleration you want to find, and <em>N₁</em> is the magnitude of the normal force exerted by the surface.

Block 2 (middle) has much the same information:

• net horiz. force

∑ <em>F</em> = <em>T₁</em> - <em>T₂</em> - <em>f₂</em> = <em>m₂a</em>

• net vert. force

∑ <em>F</em> = <em>N₂</em> - <em>m₂g</em> = 0

with similarly defined symbols.

The same goes for block 3 (leftmost):

• net horiz. force

∑ <em>F</em> = <em>T₂</em> - <em>f₃</em> = <em>m₃a</em>

• net vert. force

∑ <em>F</em> = <em>N₃</em> - <em>m₃g</em> = 0

We have <em>m₁</em> = <em>m₂</em> = <em>m₃</em> = 0.8 kg, so I'll just replace each with <em>m</em>. It follows that each normal force has the same magnitude, <em>N₁</em> = <em>N₂</em> = <em>N₃</em> = <em>mg</em>. And as a consequence of that, each frictional force has the same magnitude, <em>f₁</em> = <em>f₂</em> = <em>f₃</em> = 0.4<em>mg.</em>

In short, the relevant equations are

[1] … 57.3 N - <em>T₁</em> - 0.4<em>mg</em> = <em>ma</em>

[2] …<em>T₁</em> - <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>

[3] … <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>

<em />

Adding [1], [2] and [3] together eliminates the tension forces, and we get

57.3 N - 1.2<em>mg</em> = 3<em>ma</em>

<em />

Solve for <em>a</em> :

57.3 N - 1.2 (0.8 kg) (9.8 m/s²) = 3 (0.8 kg) <em>a</em>

57.3 N - 9.408 N = (2.4 kg) <em>a</em>

<em>a</em> = (47.892 N) / (2.4 kg)

<em>a</em> ≈ 20.0 m/s²

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