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3 years ago

When the car carrying the unbelted driver crashes, what brings the driver to a stop

1 answer:

4 0

When the driver is not wearing a seat belt and the car hits a brick wall,
the driver keeps going and doesn't stop, (because of the inertia of the
driver's body).

-- If the driver is lucky, the car is going slow enough, AND it has a
driver-side airbag, AND the airbag works, AND it inflates fast enough,
AND it cushions the driver well enough to avoid the driver's death.

-- If an airbag doesn't stop the driver, then one or more of these brings
the driver to a stop:

... the car's engine block crashing in through the dashboard
... the steering wheel
... the windshield
... the same brick wall that stopped the car.

Any one of these is probably fatal.

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What is the speed of a bobsled whose distance-time graph indicates that it traveled 124m in 26s

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It's average speed during that 26 seconds was about 4.77 m/s. Without seeing the graph, we can't tell if it was going faster or slower at any particular time during that period. All we can tell is its average for the full interval.

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4 years ago

If you do 1500 J of work hoisting a 20 kg bale of hay , to what height did you lift it

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Explanation:

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3 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

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