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Anastasy [175]
3 years ago
9

Two soccer players run toward each other. One player has a mass of 85 kg

Physics
1 answer:
sveta [45]3 years ago
8 0

Answer:

Total momentum, p = 55 kg-m/s

It is given that,

Mass of player 1, m₁ = 85 kg

Mass of player 2, m₂ = 105 kg

Speed of player 1, v₁ = -8 m/s (west)

Speed of player 2, v₂ = 7 m/s (east)

Momentum is equal to the product of mass and velocity. For this system, momentum is given by :

p=m_1v_1+m_2v_2p=m

1

v

1

+m

2

v

2

p=85\ kg\times (-8\ m/s)+105\ kg\times 7\ m/sp=85 kg×(−8 m/s)+105 kg×7 m/s

p = 55 kg-m/s

The total momentum of the system made up of the two players is 55 kg-m/s.

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The new gravitational force of the two objects will be 432 N.

The given parameters:

  • <em>Force between object 1 and object 2, F = 18 N</em>

  • <em>Let the mass of object 1 = m1</em>
  • <em>Let the mass of object 2 = m2</em>
  • <em>Let the distance between the two object = r</em>

The gravitational force between the two objects is calculated by applying Newton's law of universal gravitation;

F = \frac{Gm_1m_2}{r^2} \\\\G = \frac{Fr^2}{m_1m_2}\\\\G = \frac{Gm_1m_2}{r^2} \\\\G = \frac{18r^2}{m_1m_2}

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  • mass of the object 2 = 3m2
  • distance between the two object = ¹/₂r

\frac{18r^2}{m_1 m_2 } = \frac{F_2(\frac{1}{2}r)^2 }{2m_1 \times 3m_2} \\\\\frac{18r^2}{m_1 m_2 } = \frac{F_2 \times r^2}{4 \times 2m_1 \times 3m_2} \\\\\frac{18r^2\times 4 \times 2m_1 \times 3m_2}{m_1 m_2\times r^2 } = F_2\\\\432 \ N = F_2

Thus, the new gravitational force of the two objects will be 432 N.

Learn more about Newton's law of universal gravitation here: brainly.com/question/9373839

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