Part A In Europe, gasoline efficiency is measured in km/L. If your car\'s gas mi

Question

Part A

In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 30.0 mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions: 1km=0.6214mi and 1gal=3.78L

Part B

While in Europe, if you drive 119 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per liter and your car's gas mileage is 28.0 mi/gal ? Assume that 1euro=1.26dollars.

Part C

A sprinter set a high school record in track and field, running 200.0 m in 21.4 s . What is the average speed of the sprinter in kilometers per hour?

Part D

A specific brand of gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-g piece. How many kilograms of dietary fat are in a box containing 4.00 lb of candy?

Explanation / Answer

Part A)

we know that

1 km = 0.6214 mi

so

distance = 142 x 0.6214 mi = 88.2388 mi

now

30 = 88.2388 / volume of gasoline

volume of gasoline = 2.9412933 gal

now

1 gal = 3.78 L

so

volume of gasoline = 2.9412933 x 3.78

volume of gasoline = 11.12 L

so

11.12 L of gasoline is required

B)

total distance = 119 x 7 = 833 km

1 km = 0.6214 mi

so

total distance = 833 x 0.6214 = 517.626 mi

now

28 = 517.626 / volume

volume = 18.48665 gal

1 gal = 3.78 L

so

volume of gas = 18.48665 x 3.78

volume of gas = 69.88 L

now

cost = 1.1 x 69.88

cost = 76.87 euro

cost = 76.87 x 1.26

cost = 96.85 dollars

C)

we know that

1 hr = 60 min = 60 x 60 s

1 hr = 3600 s

speed = distance / time

speed = 200 x 10-3 km / 21.4 x (1/36000) hr

speed = 33.645 km/ hr

so

the average speed is 33.645 km / hr

D)

we know that

1 lb = 453.592 g

so

amount of candy = 4 x 453.592 = 1814.368 g

now

amount of fat = 1814.368 x 7 / 22.7

amount of fat = 559.5 g

amount of dietary fat is 0.5595 kg

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