Answer:

Explanation:
Here two charges are placed at distance "d" apart
now the net value of electric field at some position between two charges will be ZERO
so we will have
electric field due to charge 1 = electric field due to charge 2

Let the position where net field is zero will lie at distance "r" from q1

now we will have

now square root both sides

now we have

so we have

The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
Learn more about the Newton's equation of motion here:
brainly.com/question/8898885
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Answer:
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Density: g/mL, kg/cubic meter
Volume: L, teaspoon
Mass: g, MeV/sq. C