Much of the precipitation in large bodies of water occurs at the surface. The ocean loses about 37000 km cubed considering evaporation and precipitation.
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.
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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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Answer:
The slope of the graph is what you need. That tells you the speed not the velocity. In order to find the velocity you would also need to know the direction of the motion.
Answer:
Explanation:
a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²
500 = 5 x 10⁻³ k ,
k = (500/5) x 10³ = 10⁵ N/m
b )
k = 4.5 x 10¹ = 45 N/m
Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J
This energy gets dissipated by friction .
work done by friction = μ mg d
d is the distance traveled under friction
so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2
μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .
