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liberstina [14]
3 years ago
6

If a 990 kg car is traveling on the road and the Ff is 360 N to the east and the applied force is 1330 N to the west, what is th

e net force? Choose the best answer. (Remember to draw a force diagram to solve).
Physics
1 answer:
lubasha [3.4K]3 years ago
7 0

(1330N-west  +  360N-east) = (1330N-west  -  360N-west) = 970N-west .

All that business about a car, its mass, where it is, and where
it's going just gets in the way.  None of it is needed.

By the way . . . you asked us to "choose the best answer".  From what ?
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The total distance from the graph is calculated as follows;

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  • second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm
  • second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm
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average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

  • This shortest distance is the straight line connecting the start and end position. Call this line P
  • From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
  • Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of  line P.

                                                ↓end

                                                ↓

                                                ↓ 6cm

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  start -------------13 cm------------

Use Pythagoras theorem to solve for P.

P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm

The average velocity of the ant is calculated as;

average \ velocity= \frac{\Delta displacemnt  }{total\ time }= \frac{14.318 \ cm}{105 \  s} = 0.136 \ cm/s  \\\\

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