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Jan. 31, 2023•0 likes•2 views

Jan. 31, 2023•0 likes•2 views

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Problem #5) Just as a car tops a 45 meter high hill with a speed of 81 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill? B) If the slope in Problem 5 is 54 degrees, the height 14 meters and the magnitude of the work done by friction is 150 joules, what is the coefficent of friction (to 2 decimal places)? Solution Problem 5)- This question is based on the energy conservation. So, applying energy conservation here between initial situation to final situation- Total energy)i = Total energy)f KEi+ PEi = KEf + PEf (Here, at the max high KEf = 0 as velocity at there will be zero) 0.5*M*(81*5/18) 2 + M(10)(45) = M(10)*h h = 70.31 m Problem B)- Work)friction = F*(distance coverd in friction force acted) But for this equation we need the mass of the car. Please provide the mass of car. Thankyou !!! Have fun Please Rate High. .

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- 1. Problem #5) Just as a car tops a 45 meter high hill with a speed of 81 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill? B) If the slope in Problem 5 is 54 degrees, the height 14 meters and the magnitude of the work done by friction is 150 joules, what is the coefficent of friction (to 2 decimal places)? Solution Problem 5)- This question is based on the energy conservation. So, applying energy conservation here between initial situation to final situation- Total energy)i = Total energy)f KEi+ PEi = KEf + PEf (Here, at the max high KEf = 0 as velocity at there will be zero) 0.5*M*(81*5/18) 2 + M(10)(45) = M(10)*h h = 70.31 m Problem B)- Work)friction = F*(distance coverd in friction force acted) But for this equation we need the mass of the car. Please provide the mass of car. Thankyou !!! Have fun Please Rate High.