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:
Best at conductioning heat- Solid
Explanation:
<u><em>In order for heat to conduct it has to pass from particle to particle.</em></u>
A solid has particles closely compacted so the heat is able to tranfer quickly from particle to particle.
A liquid has isn't as compacted it actually has enough space to slide over each other
A gas has very far apart particles and takes heat longer to transfer from one particle to another.
Which excerpt are you talking about?
<span>Answer:
1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use.
Given: m= 1200kg v initial = 95km/hr v final = 0
2nd, focus on the units - in most cases units speak for the concept
the unit of the unknown is kcal, thus its the unit of energy or work
so, W = ?
3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present
since W = delta K.E =delta P.E
W= 0.5m( vf^2 - vi^2) ---> best formula
4th, Substitute the given to the formula
since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J
we express first 95 km/hr to m/s
95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec
W= 0.5(1200kg)[(0^2- (26.39m/sec)^2]
W=600 kg(0 - 696.43m^2/s^2)
W=600kg(-696.43m^2/s^2)
W=417859.3Nm or 417859.3 J
W = 417859.3 J x 1 cal /4.19 J
W = 99,727.7 cal or 99.728 kcal</span>
The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground. It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.
It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.
More and more PE turns into KE as the ball falls, all the way down.
When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.
