2. -4 points OSColPhya1 7.1.004. My Notes Ask Your Teache Suppose a car travels
ID: 2034386 • Letter: 2
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2. -4 points OSColPhya1 7.1.004. My Notes Ask Your Teache Suppose a car travels 108 km at a speed o 35.0 m s, and uses 2.20 gallons o 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 1.3 x 108 J per gallon.) (a) What is the force exerted to keep the car moving at constant speed? (b) I the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 mys? gallons Additional Materials Reading Submit Answer Sawe Progress Practice Another Version 3. -f4 points OSColPhys1 72012 My Notes Ask Your Teache (a) Calculate the force needed to bring a 1100 kg car to rest from a speed of 95.0 kmyh in a distance of 105 m (a fairly typical distance for a nonpanic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), e. find the ratio of the force in partb) to the force in part(a) (force in part (b) force in part (a)) Additional Materials RooingExplanation / Answer
2)Given,
d = 108 km = 1.08 x 10^5 m ; E = 1.3 x 10^8 J/gallon
v = 35 m/s ; V = 2.2 gallons
a)work done will be:
Wf = 2.2 x 1.3 x 10^8 x 0.3 = 8.58 x 10^7 J
Wf = Ff d => Ff = Wf/d
Ff = 8.58 x 10^7/1.08 x 10^5 = 794.44 N
Hence,Ff = 794.44 N
b)V = 28/30 x 2.2 = 2.05 gal
Hence, V = 2.05 Gal
3)
(a)Given,
m = 1100 kg ; v = 95 km/h = 26.4 m/s ; s = 105 m
We know from eqn of motion
v^2 = u^2 + 2 a s
a = (0 - 26.4^2)/2 x 105 = - 3.32 m/s^2
F = ma = 1100 x 3.32 = 3652 N
Hence, F = 3652 N
b)s = 2 m
a = 0^2 - 26.4^2/2 x 2 = 174.24 m/s^2
F = ma = 1100 x 174.24 = 191664 N
Fb/Fa = 191664/3652 = 52.48
Hence, Fb/Fa = 52.48
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