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Yakvenalex [24]
3 years ago
8

2. Harry is pushing a car down a level road at 2.0 m/s with a force of 243 N. The total force

Physics
1 answer:
Arte-miy333 [17]3 years ago
3 0

b) Equal to 243 N.

Explanation:

The total force acting on the car in the opposite direction including the road friction and air resistance is equal to 243 N.

This is in conformity with newton's third law of motion.

Newton's third law of motion states that "action and reaction are equal and opposite in direction. "

  • The action force is that of the pull by Harry acting on the car.
  • The reaction force is in the opposite direction.
  • Both action and reaction force equal and opposite and magnitude and direction

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

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2 years ago
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

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v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

\Delta K = 9.559\times 10^{-5}\,J

4 0
3 years ago
Read 2 more answers
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