Answer:
(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m
(b) thermal energy was generated by friction is 1.88 x J
(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N
Explanation:
given information:
m = 750 kg
initial velocity, = 110 km/h = 110 x 1000/3600 = 30.6 m/s
initial height, = 22 m
slope, θ = 2.5°
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
according to conservation-energy
EP = EK
mgh =
gh =
h =
= 47.6 m
(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
thermal energy = mgΔh
= mg (h - )
= 750 x 9.8 x (47.6 - 22)
= 188160 Joule
= 1.88 x J
(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?
f d = mgΔh
f = mgΔh / d,
where h = d sin θ, d = h/sinθ
therefore
f = (mgΔh) / (h/sinθ)
= 1.88 x /(22/sin 2.5°)
= 373 N
Answer:
Explanation:
The problem is solved using the law of conservation of energy,
So
The rock strike the water with the speed of 15.78 m/sec.
The speed by which rock hit the water is calculated by the formula
v=
v=
v=15.78 m/sec
Hence, the rock strike the water with the speed of 15.78 m/sec.
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