2011CAP Prize Exam加拿大高中物理大奖赛真题答案下载

历年CAP High School Prize Exam加拿大高中物理大奖赛

真题与答案下载

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2011 CAP Prize Exam完整版真题免费下载

共计3小时考试时间

此套试卷由23道选择题以及3道大题组成

每道大题含有不同数量的小题

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2011 CAP Prize Exam完整版真题

2011CAP Prize Exam加拿大物理奥林匹克完整版答案免费下载

部分真题答案预览：

Part A选择题部分：

1. C
2. B
3. D
4. C
5. D

Part B简答题部分

Problem 1

(a) To compare the efficiencies, we need to convert them to the same unit:

 

3.785 L/gal

USA car efficiency = 30 mpg ⇒ =7.84 L/100km

(30 mpg)×(1.609 km/mile)

European car efficiency = 7.80 L/100km

The European car is the most efficient.

(b)

0.2 × 7.8 L/100km =1.56 L/100km ⇒ European car new efficiency = 6.24 L/100km

0.2 × 30 mpg =6 mpg ⇒ USA car new efficiency = 36 mpg

3.785 L/gal

USA car new efficiency = 36 mpg ⇒ =6.53 L/100km

(36 mpg)×(1.609 km/mile)

Therefore the European car is still the most efficient, by ^6.53−6.24 ×100% = 4.65%. Note that a 20% improvement

6.24

in efficiency measured in L/100km is more important than a 20% improvement in efficiency measured in mpg.

(c) The distance traveled in each part of the trip is given by the velocity during this part, multiplied by the travel time, so the total distance traveled is:

ΔX =0.5h × 60km/h +0.5h × 120km/h +0.5h × 80km/h +0.5h × 100km/h = 180km

(d) The work done by the the drag force is the sum of the work in each part of the motion:

 

4

W =_i=1 FiΔXi = FiViΔti = 240N ×60km/h ×0.5h+540N ×120km/h ×0.5h+320N ×80km/h ×0.5h +420N ×100km/h ×0.5h = 73.4 MJ

(e) The total gasoline used is given by the sum of the amounts used in each part of the motion:

4

FiViΔti

l ==

i=1 35 MJ/L×Efficiencyi 240N×60km/h 540N×120km/h 320N×80km/h 420N×100km/h

0.5h × (+ ++)/(35 MJ/L) = 16.7 L

0.075 0.15 0.11 0.135

(f) The amount of gasoline used per kilometer travelled (denoted by lkm) for each part of the motion is then given by:

FV ΔtF 1 J/m F 1 L

== × = ×× 10⁻³

^lkm 35MJ/L×Efficiency×V Δt Efficiency 35 MJ/L Efficiency 35 km

For the different speeds of the car the lkm can be calculated:

V = 60km/h ⇒ lkm =9.14 L/100km V = 120km/h ⇒ lkm = 10.3 L/100km V = 80km/h ⇒ lkm =8.31 L/100km V = 100km/h ⇒ lkm =8.89 L/100km

Therefore gasoline is used most efficiently when the car is traveling at 80km/h .

 

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