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Lelu [443]
3 years ago
12

A 72.5-kg person is riding in a car moving at 20.5 m/s (assume this is the positive direction) when the car runs into a bridge a

butment. This problem will illustrate why the invention of the airbag dramatically improved the safety of automobiles. show answer Incorrect Answer 50% Part (a) Calculate the horizontal component of the average force, in newtons, on the person if he is stopped by a padded dashboard that compresses an average of 1.15 cm.Calculate the horizontal component of the average force, in newtons, on the person if he is stopped by an air bag that compresses an average of 13.5 cm
Physics
2 answers:
MAXImum [283]3 years ago
6 0

Explanation:

Given that,

Mass of the person, m = 72.5 kg

Initial speed of the car, u = 20.5 m/s

(a) Let F_x is the horizontal component of the average force, in newtons, on the person if he is stopped by a padded dashboard that compresses an average of 1.15 cm, x = 0.0115 m and v = 0

Using the second law of motion to find it as:

F_x=m\times a

Using the third equation of kinematics to find a.

F_x=m\times \dfrac{v^2-u^2}{2s}

F_x=-m\times \dfrac{u^2}{2s}

F_x=-72.5\times \dfrac{(20.5)^2}{2\times 0.0115}

F_x=-1.32\times 10^6\ N

(b) Here, x = 13.5 cm = 0.135 m

F_x=-m\times \dfrac{u^2}{2s}

F_x=-72.5\times \dfrac{(20.5)^2}{2\times 0.135}

F_x=-1.12\times 10^5\ N

Hence, this is the required solution.

uranmaximum [27]3 years ago
6 0

Answer:

A)  f_{dashboard} = 1324701 N

B)  f_{airbag} = 112844.90 N

Explanation:

Given data:

mass of person = 72.5 kg

velocity of person = 20.5 m/s

we know that from newton's 2nd law

f = ma

f = m \frac{v^2 - u^2}{2s}

final velocity is v = 0

f = - \frac{mu^2}{2s}

f = \frac{72.5\times 20.5^2}{2\times 1.15\times 10^{-2}}

f = 1324701 N

B)

f_{airbag} = f = \frac{72.5\times 20.5^2}{2\times 1.3.5\times 10^{-2}}

f_{airbag} = 112844.90 N

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A) The change in internal chemical energy is 1.15\cdot 10^7 J

B) The time needed is 1 minute

Explanation:

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v = 8.0 m/s is the velocity

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P=\frac{E}{t}

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t = 30 min \cdot 60 = 1800 s is the time elapsed

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From the previous part, we found that in a time of

t = 30 min

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So we can setup the following proportion:

\frac{t}{E}=\frac{t'}{E'}

And solving for t',

t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min

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