Calculating Force Required to Overcome Resistance of Road & Wind at 75.0km/hr

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At a speed of 75.0 km/hr, 63.0% of a car's engine output power is used to overcome road and wind resistance. For an engine with an output of 69.00 HP, this translates to approximately 32,428.62 Watts dedicated to overcoming resistance. The formula for power, which is power equals force times distance over time, requires the speed to be converted to SI units (meters per second) for accurate calculations. Without this conversion, the resulting force will not be in Newtons. Proper unit conversion is essential for determining the correct force required to overcome resistance.
jacksondwrd
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You are testing a new car in a wind-tunnel for road and wind resistance. At a speed of 75.0 Km/hr you have found that 63.0 % of the total output power of an automobile engine is used in overcoming the resistance of the road and wind against the movement of the car. If the output power of a particular car engine is 69.00 HP, what is this resistance (ie. what is the force)? Give your answer in Newtons. (Take one HP as 746 Watts.


so i was like calcualting it like this

69HP=51474
and 60% of 51474 is 32428.62

then i did the formula of the power

power=forcexdistance/time

and got that answer but it is wrong

can anyone explain to me what i did wrong here
 
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I am guessing you did not convert your 75 km/hr in SI units m/s. If you didn't, you won't get your answer in Newtons.
 

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