Suppose a car travels 108 km at a speed of 40.0 m/s, and uses 1.80 gallons of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (The energy content of gasoline is 1.30 ✕ 108 J per gallon.) (a)What is the force (in N) exerted to keep the car moving at constant speed? _____ N   (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?

Q: Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the…

A: Given,

Q: A 53.0 kg cheetah accelerates from rest to its top speed of 33.3 m/s. (a) How much net work (in J)…

A: a)The required work done is,

Q: Suppose a car travels 108 km at a speed of 35.0 m/s, and uses 1.80 gallons of gasoline. Only 30% of…

A: (a). The amount of work done to keep the car moving at constant speed can be calculated as follows,

Q: A mechanic pushes a 2280 kg car from rest to a speed of v, doing 5252 J of work in the process.…

A:

Q: A 63.0 kg cheetah accelerates from rest to its top speed of 31.5 m/s. HINT (a) How much net work (in…

A: (a) Using Work-Energy theorem, net work required is,

Q: Suppose a car travels 108 km at a speed of 40.0 m/s, and uses 2.20 gallons of gasoline. Only 30% of…

A: a) The required force is,

Q: Fig. 4b shows a spring loaded along an incline plane. When s = 0.6 m, the spring is unstretched and…

A: Force is a vector quantity that means it has both magnitude and direction to it. According to…

Q: Suppose a car travels 108 km at a speed of 20.0 m/s, and uses 2.10 gallons of gasoline. Only 30% of…

A: Given : d = 108 km = 108 x 103 m v = 20 m/s V = 2.10 gallons E = 1.30 x 108 J per gallon e = 30% =…

Q: Suppose a car travels 108 km at a speed of 35.0 m/s, and uses 2.20 gallons of gasoline. Only 30% of…

A: Given, The energy content of gasoline is 1.30 ✕ 108 J per gallon a) F=? Volume, V=2.2…

Q: An object of mass (m = 5.4 kg) moves under the influence of one force. That force causes the object…

A:

Q: Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the…

A: The work done by engine of the car can be obtained as

Q: uppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.00 gallons of gasoline. Only 30% of…

A:

Q: A 61.0 kg cheetah accelerates from rest to its top speed of 32.5 m/s. HINT (a) How much net work (in…

A:

Q: Suppose a car travels 108 km at a speed of 25.0 m/s, and uses 1.80 gallons of gasoline. Only 30% of…

A:

Q: Suppose a car travels 108 km at a speed of 40.0 m/s, and uses 2.10 gallons of gasoline. Only 30% of…

A: The energy content of gasoline,

Q: Suppose a car travels 108 km at a speed of 20.0 m/s, and uses 1.80 gallons of gasoline. Only 30% of…

A: The work done by a force on moving a body by a distance is given by W = F.s, where F = the force…

Q: Suppose a car travels 108 km at a speed of 40.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of…

A:

Q: Suppose a car travels 108 km at a speed of 20.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of…

A: Give data, Distance covered by the car, S = 108 km Initial velocity of the car, u = 20.0 m/s Total…

Q: Suppose a car travels 108 km at a speed of 20.0 m/s, and uses 2.00 gallons of gasoline. Only 30% of…

A: Given data: Distance d = 108 km Speed v1 = 20 m/s Used gasoline = 2.00 gallons % of gasoline used…

Q: Problem 1: A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well…

A: Given data:  Work done due to force F1 Wf = 484.5 J Work done due gravity Force Fg Wg = -344.8 J…

Q: Suppose a car travels 108 km at a speed of 25.0 m/s, and uses 2.20 gallons of gasoline. Only 30% of…

A: Given : d = 108 km = 108 x 103 m v = 25 m/s V = 2.20 gallons e = 30% = 0.30 E = 1.3 x 108 J per…

Q: A 64.0 kg cheetah accelerates from rest to its top speed of 30.7 m/s. HINT (a) How much net work (in…

A:

Q: A 64.0 kg cheetah accelerates from rest to its top speed of 32.5 m/s. (a) How much net work (in J)…

A: (a) According to the work energy theorem, W=∆K=12mv2-12mu2=12mv2-u2 Substitute the values, to find…

Q: Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.20 gallons of gasoline. Only 30% of…

A: (A)The number of gallons of gasoline is 2.20.The percentage of gasoline goes into useful work is…

Q: Suppose a car travels 108 km at a speed of 35.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of…

A: (a) Force from the expression for work done W=FdF=Wd=30%1.90 gal1.3×108 Jgal 108×103 m=686 N

Q: a) A stationary bowler is getting ready to throw the bowling ball. She holds a 60 kg bowling ball…

A: Work energy theorem: Work done (W) = Change in kinetic energy. By law of conservation of energy:…

Q: The spring-flex exercise system consists of a spring with one end fixed and a handle on the other…

A:

Q: Suppose a car travels 108 km at a speed of 35.0 m/s, and uses 2.10 gallons of gasoline. Only 30% of…

A: The required energy is, E=30%1.3×108 J/gallon2.10 gallon=8.19×107 J The distance traveled by car is,…

Q: Mrs. Salcedo's car of mass 2.4 x 10 kg travelling on the highway when the teacher receives an…

A:

Q: A child pulls a sled (mass 8.00 kg ) over a distance of 20.0 m [West] by applying a force of 30.0 N…

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Q: A 62.0 - kg cheetah accelerates from rest to its top speed of32.0 m/s. (a) How much net work is…

A: (a)According to work energy theorem the net work cone by the cheetah to reach the top speed is.

Q: A net force acts on an object and accelerates it from rest to a speed v1. In doing so, the force…

A: