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3 years ago

The driver and front seat passengers should sit at least _____ inches away from the air bag.

1 answer:

3 0

Air bags are kept in the steering wheel or dashboard and expand during a serious collision, usually a front collision that occurs at over 10 mph. To do its important job, an air bag comes out of the dashboard at up to 200 mph, faster than the blink of an eye. It is estimated around 10 inches of space to inflate. The force of an air bag can hurt those who are too close to it. Driver and front seat passengers should be moved as far back as practical, mainly people of short built. It is highly suggested that you be seated no less than 10 inches away from the air bag. Therefore, the answer is letter b.

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Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

leonid [27]

Answer:

$KE_A=33\ J$

$KE_B=99\ J$

Explanation:

Given:

Let mass of the particle B be, $m_B=m$

then the mass of particle A, $m_A=3m$

Energy stored in the compressed spring, $E=132\ J$

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

Using the conservation of linear momentum:

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

Now the kinetic energy of particle B:

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

$KE_B=99\ J$

Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

Now the kinetic energy of particle A:

$KE_A=\frac{1}{2}m_A.v_A^2$

$KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$

$KE_A=33\ J$

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Elden [556K]

Answer:

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