Question

Son) A 150 HP electric vehicle with 72 KWh 800 V battery cruizes on the highway at a constant 85 km/hr, which is the most efficient speed for that brand. Its fully charged batteries last 7.6 hours at this constant speed. a) What is the power consumed at this speed? (50 pts), b) What is the current drawn ? (50 pts)

          Son) A 150 HP electric vehicle with 72 KWh 800 V battery cruizes on the highway at a constant 85 km/hr, which is the most efficient speed for that brand. Its fully charged batteries last 7.6 hours at this constant speed.
a) What is the power consumed at this speed? (50 pts),
b) What is the current drawn ? (50 pts)
        
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Son) A 150 HP electric vehicle with 72 KWh 800 V battery cruizes on the highway at a constant 85 km/hr, which is the most efficient speed for that brand. Its fully charged batteries last 7.6 hours at this constant speed.
a) What is the power consumed at this speed? (50 pts),
b) What is the current drawn ? (50 pts)

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A 150 HP electric vehicle with a 72 kWh 800V battery cruises on the highway at a constant 85 km/hr, which is the most efficient speed for that brand. Its fully charged batteries last 7.6 hours at this constant speed. b) What is the current drawn? 50 pts
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Transcript

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00:01 In this problem, we have been given that there is a car and the battery of this car has a capacity of 100 kilowatt hours.
00:11 Basically this is the energy and we know that one kilowatt hour that's 3 .6 times 10 raised to 6 jules so 100 kilowatt art that will be 3 .6 times 10 raise to 6 times 100 jules and we are also given that a battery consumes a point one seven kilowatt hour of energy on driving through one kilometer so we can see point one seven kilowatt hour per kilometers that is consumed and here this car is being driven at a speed of 100 kilometers per hour and we are required to determine the power output of this battery provided that the car maintains the speed so we will use this expression of the power but before that let's compute the time here so as we are given that point one seven kilowatt hour of energy that's consumed every one kilometer so one kilowatt hour of energy will be consumed in one over point one seven kilometers and we have the total capacity to be consumed as 100 kilowatt hour so we multiply both sides with 100 and this gets us the distance traveled as 100 over point one seven kilometers so this is the distance and we know that the speed that's given to us that's hundred that's equal to the distance over the time and from here we solve so we get the time coming out to be this hundred gets cancelled and time comes out to be one by point one seven that will be hundred over 17 hours and now we can compute the power output of this battery.
02:07 So this power output, that will be the energy.
02:09 And let's take the energy in jules.
02:11 So it will be 3 .6 times 10 raise to 6 times 100.
02:17 And this is in jules divided by the time.
02:21 And let's even convert this time in seconds because then only we'll be getting the power in terms of what.
02:30 So we know that one hour...
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