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dezoksy [38]
3 years ago
9

A test car is driving toward a solid crash-test barrier with a speed of 46 mi/h. Two seconds prior to impact, the car begins to

brake, but it is still moving when it hits the wall. After the collision with the wall, the car crumples somewhat and comes to a complete stop. In order to estimate the average force exerted by the wall on the car, what information would you need to collect?
Physics
1 answer:
Mekhanik [1.2K]3 years ago
4 0

Answer:

we have to measure distances and time, possibly with an automated system since the values ​​are very small

Explanation:

For this exercise we can use the relationship between momentum and momentum

        I = ∫ F dt = Δp

        F t = m v_{f} - mv₀

In the exercise they indicate that the final speed is zero

         F t = - m v₀

         F = -m v₀ / t

With this equation we can find what measurements should be carried out.

To find the speed with which the car collides with the wall, less measure the displacement and its time during the braking process before reaching the wall and from here find the speed with which it reaches the wall.

During the impact, we must find the time that the vehicle is in contact with the wall in the first approach is equal to the time that the car takes to reach the final speed of zero.

In summary we have to measure distances and time, possibly with an automated system since the values ​​are very small

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artcher [175]

Answer:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J),  The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

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v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

<em>Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)</em>

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

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<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

4 0
3 years ago
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sin\theta_c=\frac{n_2}{n_1}

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By taking n_2 and replacing we obtain:

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hope this helps!!

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