A small model car with mass m travels at a constant speed on the inside of a track that is a vertical circle radius 5.00 m. If the normal force exerted by the track on the car when it is at the bottom of the track (point A) is equal to 2.50 mg, how much time does it take the car to complete one revolution around the track?

Based on the information given, it can be deduced that the time taken will be 3.63 seconds to complete the revolution.

It should be noted that when the car is at the bottom, the formula to use from Newton's equation will be:

N - mg = (m × v²) / r

2.5 - 1 = v² / (g × 5)

1.5 = v² / (10 × 5)

1.5 = v² / 50

v² = 50 × 1.5

v = ✓75

v = 8.66m/s

Therefore, the time taken will be:

= (2 × pi × r) / v

= (2 × 3.142 × 5) / 8.66

= 3.63 seconds.

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increases it

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information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!