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?
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This is the same question as the one previously but with more details, so I will just use my previous answer.
1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.
So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
They seem to cancel each other out which is odd
<span>The Dynamo Theory
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The distance between the two charges is 
Explanation:
The electrostatic force between two charged objects is given by Coulomb's law:
where:
is the Coulomb's constant
are the charges of the two objects
r is the separation between the two charges
In this problem, we are given the following:
Therefore, we can rearrange the equation to solve for r, the distance between the two charges:
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