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:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
Answer:
Newton's first law
Explanation:
Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. Therefore, when the ice is smooth, friction gets lesser, and the force acted on that Puck will be decreased.
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