How many kilocalories are generated when the brakes are used to bring a 1200-kg car to rest from a speed of 95 km/h ? 1 kcal = 4186 j?

1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use. Given: m= 1200kg v initial = 95km/hr v final = 0 2nd, focus on the units - in most cases units speak for the concept the unit of the unknown is kcal, thus its the unit of energy or work so, W = ? 3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present since W = delta K.E =delta P.E W= 0.5m( vf^2 - vi^2) ---> best formula 4th, Substitute the given to the formula since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J we express first 95 km/hr to m/s 95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec W= 0.5(1200kg)[(0^2- (26.39m/sec)^2] W=600 kg(0 - 696.43m^2/s^2) W=600kg(-696.43m^2/s^2) W=417859.3Nm or 417859.3 J W = 417859.3 J x 1 cal /4.19 J W = 99,727.7 cal or 99.728 kcal

Question was not very clear if there are specific answer choices tell me and i'll try to you more. hope this though2 laps is 1 min. we know that for each min she does two times the amount of laps. for 12 laps we can do 12/2 which does give us 12 laps in 6 mins not 16 mins.

the simple machine has three main features. they are:

loadfulcrumeffort

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever