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A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force between the tires and the road is 7100 N. How far will the car slide once the brakes are applied? ...?

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vo = 25 m/sec 
vf = 0 m/sec 
Fμ = 7100 N (Force due to friction) 
Fg = 14700 N 
With the force due to gravity, you can find the mass of the car: 
F = ma 
14700 N = m (9.8 m/sec²) 
m = 1500 kg 
Now, we can use the equation again to find the deacceleration due to friction: 
F = ma 
7100 N = (1500 kg) a 
a = 4.73333333333 m/sec² 
And now, we can use a velocity formula to find the distance traveled: 
vf² = vo² + 2a∆d 
0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d 
0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d 
-625 m²/sec² = (-9.466666666667 m/sec²) ∆d 
∆d = 66.0211267605634 m 
∆d = 66.02 m
pstnonsonjoku

From the calculations and the data supplied, the braking distance is 65 m

What is the braking force?

The braking force is the force that is applied as the object comes to rest.

We know that the mass of the car is obtained from; 14,700 N/10 m/s^2 = 1470 Kg

Now;

a = F/m = 7100 N/1470 Kg

a = 4.8 m/s^2

Given that;

v^2 = u^2 - 2as

u^2 = 2as

s =  u^2/2a

s = (25)^2/2 * 4.8

s = 625/9.6

s = 65 m

Learn more about braking distance:https://brainly.com/question/13666022

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