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mina [271]
3 years ago
15

Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by a distance of 1 meter.

Physics
1 answer:
Makovka662 [10]3 years ago
6 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The Gravitational Force between given objects will be ~

  • 9.34 \times  {10}^{ - 6}  \:  \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where ~

  • F = gravitational force

  • m_1 = mass of 1st object = 70 kg

  • m_2 = mass of 2nd object = 2000 kg

  • G = gravitational constant = 6.674 × {10}^ {-11}

  • r = distance between the objects = 1 m

Let's calculate the force ~

  • F =  \dfrac{6.674 \times 10 {}^{ - 11} \times 70 \times 2000 }{1 {}^{2} }

  • F =6.674 \times 7 \times 2 \times 10 { }^{ - 11} \times 10 {}^{4}

  • 93.436 \times 10 {}^{ - 7}

  • 9.3436  \times 10 {}^{ - 6} \:  \:  newtons
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A man's higher initial acceleration means that a man can outrun a horse over a very short race. A simple - but plausible - model
Lera25 [3.4K]

Answer:

 t_man = 10.16 s,   t_horse = 10.73 s,  the winner is the man

Explanation:

To solve this problem we are going to find the time of each one separately.

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         x₁ =  v₀ t₁ + ½ a₁ t₁²

as it comes out of rest its initial velocity is zero

        x₁ = ½ a₁ t₁²

        x₁ = ½ 6.0 1.8²

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at this point its speed is

        v₁ = v₀ + a t

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        v₁ = 10.8 m / s

From here on it continues at constant speed, the distance that the man needs to travel from the point where the man leaves at 100m is

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        x₂ = 100- 9.72

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the time for this part is

        v₁ = x₂ / t₂

         t₂ = x₂ / v₂

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        t_man = t₁ + t₂

        t _man = 1.8 + 8.36

        t_man = 10.16 s

We repeat the calculation for the horse

distance traveled during the acceleration period

         x₃ = v₀ t + ½ a₂ t₃²

as part of rest its initial velocity is zero

        x₃ = ½  a₂ t₃²

        x₃ = ½  5.0  4.8²

        x₃ = 57.6 m

the velocity at this point is

         v₃ = v₀ + a₂ t₃

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the rest of the route is at constant speed, the remaining distance

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the time to go through it is

         t₄ = x₄ / v₃

         t₄ = 142.4 / 24

         t₄ = 5.93 s

the total time for the horse is

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         t_horse = 10.73 s

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