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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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Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

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Explanation:

happy to help:)

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3 years ago
Under what circumstances will the distance traveled by an object be the same as the magnitude of the displacement of an object?​
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Kūna veikia jėga kurios momentas 0,4 N × m petys - 5 cm. Koks šios jėgos dydis?​
umka21 [38]

Answer:

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Explanation:

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We need to find the magnitude of this force. We know that, the torque acting on an object is given by :

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Answer:

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