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mihalych1998 [28]
3 years ago
9

Learning Goal:

Physics
1 answer:
Elina [12.6K]3 years ago
3 0

Answer:

A)   k = 2,858 10⁴ N / m ,  B)  x_total = - 5,812 10⁻¹ m

C) it is very possible that the obtained value is not realistic for small vehicles

Explanation:

Part A

In this case we can use Hooke's law

          F = - k x

          force is the weight of the driver

          F = W

          mg = - k x

          k = - mg / x

        the springs are compressed x = - 2,4 10⁻² m

         k = - 70 9.8 / (-2.4 10⁻²)

         k = 2,858 10⁴ N / m

Part B

Since we have the spring constant we must find the complete weight, for this we look for the total masses

   

each mass is the number of element by the mass of an element

     M = 23 70 + 3 15 + 5 3 + 1 25

     M = 1695 kg

     F = -k x

    F = W = M g

    Mg = - k x_total

   x_total = -M g / k

   x_total = -1695 9.8 / 2,858 10⁴

   x_total = - 5,812 10⁻¹ m

The negative sign indicates that the springs are compressing

Part C

The truck has lowered 0.58 m = 58 cm

This drop is very large probably in a real vehicle with this drop it would be touching the ground or very close, therefore it is very possible that the obtained value is not realistic for small vehicles

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What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
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Answer:

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Explanation:

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a = \frac{v^{2}}{r}  (1)

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v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

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r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

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Then, equation 2 can be used

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T = 24h . \frac{3600s}{1h} ⇒ 84600s

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