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Volgvan
4 years ago
6

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

peed) brakes and begins to slow down with the largest deceleration possible without skid- ding. Considering an average human reaction, you press your brakes 0.507 s later. You also brake and decelerate as rapidly as possible without skidding. Assume that the coefficient of static friction is 0.868 between both cars’ wheels and the road. The acceleration of gravity is 9.8 m/s2 . Calculate the acceleration of the car in front of you when it brakes. Answer in units of m/s2.
Physics
1 answer:
densk [106]4 years ago
4 0

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

F_f = \mu m g

So the acceleration of car due to friction is given as

F_{net} = - \mu mg

a = \frac{F_{net}}{m}

a = -\mu g

now it is given that

\mu = 0.868

g = 9.81 m/s^2

so here we have

a = -0.868 * 9.81

a = -8.52 m/s^2

so the car will accelerate due to brakes by a = - 8.52 m/s^2

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3 years ago
Which EM wave is about the size of humans?
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3 years ago
Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau
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The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

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From Columb's law;

\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9}  = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\  q = 7 \times 10^{-13} \ C

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brainly.com/question/1616890

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7 0
2 years ago
Jackson throws a football 30 meters at a speed of 15 m/s. How long was the football in the air before Laurence caught it for tou
ch4aika [34]

Answer:

2s

Explanation:

Given parameters:

Distance  = 30m

Speed  = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

    Speed  = \frac{disance }{time}  

 Time taken  = \frac{distance }{speed }   = \frac{30}{15}  = 2s

The time it takes is 2s

6 0
3 years ago
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