7.80\Xd710^7/1.0\Xd710^4 Scientific Notation.
7.80×10^7/1.0×10^4 scientific notation.
ANSWER : 7.8 × 10^3STEP BY STEP EXPLANATION :
(7.80 × 10^7)/(1.0 × 10^4) = 7.8 × 10^3
OK.
O Avoid An Accident A Driver Steps On The Brakes To Stop A 950 Kg Car Traveling At 60 Km/H. If The Braking Is Distance Is 25 M, How Much Force Is Need
o avoid an accident a driver steps on the brakes to stop a 950 kg car traveling at 60 km/h. if the braking is distance is 25 m, how much force is needed to stop the car Good Day... Problem: To avoid an accident a driver steps on the brakes to stop a 950 kg car traveling at 60 km/h. if the braking is distance is 25 m, how much force is needed to stop the car. Given Data: Velocity initial = 60 km/hr 60 km/hr × 1000 m/ 1km × 1 hr / 3600 s = 16.667 m/s Velocity final = 0 m/s (stop) d (distance) = 25 meters a (acceleration) = ? unknown Solution: First: We will solve for acceleration with the formula Vf^2 = Vi^2 + 2 ad To solve for a (acceleration) we can use a = Vf^2 - Vi^2 / 2d a = Vf^2 - Vi^2 / 2d = (0 m/s)^2 - (16.667 m/s)^2 / 2 × 25 m = - 277.789 m^2/s^2 / 50 m = -5.556 m/s^2 (decelerating) Second: The force needed to stop the car is the mass times acceleration: m (mass) = 950 kg a (acceleration) = -5.556 m/s^2 F = ma = 950 kg × -5.556 m/s^2 = - 944.44 kg.m...
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