John\'s car weighs 1.3 tons (= U.S. tons or short tons). He adds an air conditio
ID: 112112 • Letter: J
Question
John's car weighs 1.3 tons (= U.S. tons or short tons). He adds an air conditioner to it that weighs 13 pounds. What is the total weight of the car in kg. Answer: ____ kg 1.5 inches of rain fell on a watershed that was 11 km^2 in size. How many gallons of rain fell on the entire watershed? Answer, using scientific notation: ____ gal. A small diesel engine uses 0.3 gal/hour running at a constant rpm. How many hours can the engine run on liters of fuel? Answer: ____. A small town has 3045 people, and an area of 126.4 sq. miles. What is the density of the population (i.e. people /km^2)? Answer: _____ people/km^2 An object moves at the speed of 24 cm/sec. How many seconds will it take to cover 1.05 m? Answer: _____ sec. This Workbook Assignment has 27 questions, and requires 32 answers. In order to get 95 % of the answers correct in order to get credit for this exercise, how many answers will you have to answer correctly? _____ answers.Explanation / Answer
22. John's car weight is 1.3 tons (1ton =1000kg hence 1.3ton=1300kg). Now, he added 13pounds air conditioner(1pound= 0.45359237kg hence 13pounds = 5.89670081kg).
Total weight = John's car weight(in kg) + air conditioner(in kg)
=1300+5.89670081
=1305.8967kg
23. 1.5 inches of rain fell on watershed( let us convert it in km. So, 1 inch = 2.54 x 10-5 km hence 1.5 inch = 3.988x 10-5km).
Now, hence total km3 of rain fed into area will be
3.988 x 10-5 km x 11km2= 43.868 x 10-5km3.
Now, 1km3= 2.642 x 1011gallons
hence 43.868km3= 1.159 x 1013x 10-5gallons
= 1.159 x 108 gallons of rain fed on entire watershed.
24. For one hour it requires 0.3gal of fuel( 1gal =3.785 liter, hence 0.3 gal = 1.13562 liter)
Now, 34 liters of fuel, the engine will run for
or
1.13562lit--------1hour
34lit-------------------?
By cross multiplication,
=(34 x 1)/1.13562
=29.9395 hour
=~30 hour engine will run
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