Think it would be c, 5.1 g
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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<span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365
Fsf = Static frictional force = (coefficient of static friction) * (Normal force)
So the least for you could exert to move it is equal to the Fsf.
Fsf = (0.49)(1365N)</span><span>
</span>
Pressure
= Force/Area
Area = π(d^2)/4
= π(0.4^2)/4
=0.126 m2
Pressure
= 50/0.126
= 396.825 Pa
