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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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Answer:

Explanation:

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Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%
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6.8370869499\times 10^{20}\ N

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e = Charge of electron = 1.6\times 10^{-19}\ C

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n=\dfrac{mN_AZe}{M}

q=imbalance\times \dfrac{mN_AZe}{M}

q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C

q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C

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F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N

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