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:
Friction is useful in some cases like walking and cycling ..
but it is unwanted in machines as it create unwanted sounds and heat .,due to which we loss energy
Explanation:
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Answer:
Nuclear fusion and gravitational contraction
Explanation:
In stars, there is an equilibrium between two forces, the force of gravity in the inward direction due to their own mass, and the radiation pressure in the upward direction as a consequence of the nuclear reaction in their core, that is known as hydrostatic equilibrium.
The radiation pressure is gotten from the nuclear reactions at the core (when lighter elements fuse into heavier elements), but if the nuclear reactions stop, hence, the radiation pressure will also do it and the force of gravity will overcome and break the equilibrium.
Both of that energy sources help to maintain a star's internal thermal pressure, since the contractions of the superficial layers will increase the density at the core.
Answer:
C
Explanation:
KE=1/2mv^2
KE=1/2(20)(0.1)^2
0.1^2= 0.01
KE= 1/2(20)(0.01)
KE= 0.1 J