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slavikrds [6]
3 years ago
14

In a 50 km/h head-on crash test, the steering column of passenger car 1 moved 3 cm upward and 2 cm rearward. The steering column

of passenger car 2 moved 6 cm up and 24 cm toward the rear of the car. Which of the two cars would protect people better in a crash? Explain your answer.
Physics
1 answer:
Softa [21]3 years ago
4 0

Answer:

Car 1

Explanation:

The steering column which moves the least is less likely to to the driver's chest ordinarily. Driver tends to remain in motion until restrained. Assuming a  seat belt not airbag

Generally one would compute a vector find direction and distance.  This is like solving for a hypotenuse  / in a right angled triangle problem. On face value the column moving the least is safer. The 6/24 would hit the upper chest, face, or possibly break the neck.

hence, car 1 moved 3 cm upward and 2 cm rearward is safer.

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Answer:

a

     l_o  =52 \  m

b

      l = 37.13 \ LY

Explanation:

From the question we are told that

    The  speed of the spaceship is  v  =  0.800c

    Here  c is the speed of light with value  c =  3.0*10^{8} \ m/s

    The  length is  l = 31.2 \  m

     The  distance of the star for earth is d = 145 \  light \  years

     The  speed is v_s = 2.90 *10^{8}

     

Generally the from the length contraction equation we have that

       l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}

Now the when at rest the length is  l_o

So  

      l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }

      l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }

      l_o=52 \  m

Considering b  

  Applying above equation

            l  =l_o \sqrt{1 -  [\frac{v}{c } ]}

Here l_o  =145 \  LY(light \ years )

So

           l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }

            l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }

            l = 37.13 \ LY

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3 years ago
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