Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 25, Problem 67P

(a)

To determine

Minimum power delivered by the electric motor

(a)

Expert Solution
Check Mark

Answer to Problem 67P

  26640 W

Explanation of Solution

Given:

Average force due to air drag and rolling friction

  =F=1.20 kN = 1.20 (1000) N = 1200 N

Speed of the car

  =v=80 kmh=80kmh(1000 m1km)(1h3600 s)=22.2ms

Minimum power delivered by the battery =Pbattery

Formula Used:

Power in terms of Voltage “F” and constant velocity“v” is given as

  P=Fv

Calculation:

Minimum power delivered by the battery is:

  Pbattery=FvPbattery=(1200)(22.2)Pbattery=26640 W

Conclusion:

Hence, minimum power delivered by the battery is 26640 W .

(b)

To determine

Total charge delivered by series combination of ten batteries.

(b)

Expert Solution
Check Mark

Answer to Problem 67P

  5.76×105 C

Explanation of Solution

Given:

Charge delivered by each battery =Q=160 A.h = 160(3600) A.s = 5.76×105 C

Total charge delivered by series combination of batteries =Qs

Formula Used:

Charge in terms of current “I” and “t” is given as

  Q=It

Calculation:

Power delivered by single battery is:

  Q=160 AQ= 160(3600) AQ = 5.76×105 C

In series charge remains same. Hence the charge for series combination of ten batteries is same as the charge delivered by a single battery.

Hence

  Qs= Q = 5.76×105 C

Conclusion:

Hence, thecharge delivered by series combinationis 5.76×105 C .

(c)

To determine

Total energy delivered by series combination of ten batteries.

(c)

Expert Solution
Check Mark

Answer to Problem 67P

  6.91×107 J

Explanation of Solution

Given:

Charge delivered by each battery =Q=160 A.h = 160(3600) A.s = 5.76×105 C

EMF of each battery =E=12 volts

Number of batteries in series =n=10

Total EMF provided by series combination of batteries =Es

Total energy delivered by the combination of batteries =Us

Formula Used:

Energy provided by a battery of EMF “V” by delivering a charge “Q” is given as

  U=QE

Calculation:

Total EMF provided by series combination of batteries:

  Es=nEEs=(10)(12)Es=120 volts

Total energy delivered by the combination of batteries:

  Us=QEsUs=(5.76×105)(120)Us=6.91×107 J

Conclusion:

Hence, thetotal energy delivered by the combination of batteriesis 6.91×107 J .

(d)

To determine

The distance traveled by the car before recharge is required.

(d)

Expert Solution
Check Mark

Answer to Problem 67P

  5.76×104 m

Explanation of Solution

Given:

Distance traveled by car =d

Work done by force on the car =W

Total energy delivered by the combination of batteries =Us=6.91×107 J

Formula Used:

Work done is given as

  W=Fd

Where, F is the applied force and d is the distance.

Calculation:

Work done by force on the car is given as

  W=Fd

Using conservation of energy:

  W=UsFd=6.91×107d(1200)=6.91×107d=5.76×104 m

Conclusion:

Hence, thedistance traveled by caris 5.76×104 m .

(e)

To determine

The cost per kilometer

(e)

Expert Solution
Check Mark

Answer to Problem 67P

  $0.03 km1

Explanation of Solution

Given:

Charge delivered by each battery =Q=160 A.h 

Total EMF provided by series combination of batteries =Es=120 volts

Cost per kilowatt-hour =c=9centskWh

Total cost =C

Total distance traveled =d=5.76×104 m

Total energy delivered by the combination of batteries =Us

Formula Used:

Energy provided by a battery of EMF “V” by delivering a charge “Q” is given as

  U=QE

Total cost is given as

  Total cost = (Cost per kilowatt-hour)(total energy)

Cost per unit distance is given as

  Cost per kilo-meter =Total Costdistance

Calculation:

Total energy delivered by the combination of batteries is given as

  Us=QEsUs=(160)(120)Us=19200 kWh

Total cost is given as

  C = cUsC=(9)(19200)C=172800 centsC=$1728.00 

Cost per unit distance is given as

  Cost per kilo-meter =Total CostdistanceCost per kilo-meter =$17285.76×104Cost per kilo-meter =$0.03 km1

Conclusion:

Hence total cost per kilometer comes out to be $0.03 km1

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Chapter 25 Solutions

Physics for Scientists and Engineers

Ch. 25 - Prob. 11PCh. 25 - Prob. 12PCh. 25 - Prob. 13PCh. 25 - Prob. 14PCh. 25 - Prob. 15PCh. 25 - Prob. 16PCh. 25 - Prob. 17PCh. 25 - Prob. 18PCh. 25 - Prob. 19PCh. 25 - Prob. 20PCh. 25 - Prob. 21PCh. 25 - Prob. 22PCh. 25 - Prob. 23PCh. 25 - Prob. 24PCh. 25 - Prob. 25PCh. 25 - Prob. 26PCh. 25 - Prob. 27PCh. 25 - Prob. 28PCh. 25 - Prob. 29PCh. 25 - Prob. 30PCh. 25 - Prob. 31PCh. 25 - Prob. 32PCh. 25 - Prob. 33PCh. 25 - Prob. 34PCh. 25 - Prob. 35PCh. 25 - Prob. 36PCh. 25 - Prob. 37PCh. 25 - Prob. 38PCh. 25 - Prob. 39PCh. 25 - Prob. 40PCh. 25 - Prob. 41PCh. 25 - Prob. 42PCh. 25 - Prob. 43PCh. 25 - Prob. 44PCh. 25 - Prob. 45PCh. 25 - Prob. 46PCh. 25 - Prob. 47PCh. 25 - Prob. 48PCh. 25 - Prob. 49PCh. 25 - Prob. 50PCh. 25 - Prob. 51PCh. 25 - Prob. 52PCh. 25 - Prob. 53PCh. 25 - Prob. 54PCh. 25 - Prob. 55PCh. 25 - Prob. 56PCh. 25 - Prob. 57PCh. 25 - Prob. 58PCh. 25 - Prob. 59PCh. 25 - Prob. 60PCh. 25 - Prob. 61PCh. 25 - Prob. 62PCh. 25 - Prob. 63PCh. 25 - Prob. 64PCh. 25 - Prob. 65PCh. 25 - Prob. 66PCh. 25 - Prob. 67PCh. 25 - Prob. 68PCh. 25 - Prob. 69PCh. 25 - Prob. 70PCh. 25 - Prob. 71PCh. 25 - Prob. 72PCh. 25 - Prob. 73PCh. 25 - Prob. 74PCh. 25 - Prob. 75PCh. 25 - Prob. 76PCh. 25 - Prob. 77PCh. 25 - Prob. 78PCh. 25 - Prob. 79PCh. 25 - Prob. 80PCh. 25 - Prob. 81PCh. 25 - Prob. 82PCh. 25 - Prob. 83PCh. 25 - Prob. 84PCh. 25 - Prob. 85PCh. 25 - Prob. 86PCh. 25 - Prob. 87PCh. 25 - Prob. 88PCh. 25 - Prob. 89PCh. 25 - Prob. 90PCh. 25 - Prob. 91PCh. 25 - Prob. 92PCh. 25 - Prob. 93PCh. 25 - Prob. 94PCh. 25 - Prob. 95PCh. 25 - Prob. 96PCh. 25 - Prob. 97PCh. 25 - Prob. 98PCh. 25 - Prob. 99PCh. 25 - Prob. 100PCh. 25 - Prob. 101PCh. 25 - Prob. 102PCh. 25 - Prob. 103PCh. 25 - Prob. 104PCh. 25 - Prob. 105PCh. 25 - Prob. 106PCh. 25 - Prob. 107PCh. 25 - Prob. 108PCh. 25 - Prob. 109PCh. 25 - Prob. 110PCh. 25 - Prob. 111PCh. 25 - Prob. 112PCh. 25 - Prob. 113PCh. 25 - Prob. 114PCh. 25 - Prob. 115PCh. 25 - Prob. 116PCh. 25 - Prob. 117PCh. 25 - Prob. 118PCh. 25 - Prob. 119PCh. 25 - Prob. 120P
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