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mihalych1998 [28]
3 years ago
9

Un auto de 1000 kg que lleva una velocidad de 72 km/h se detiene en una distancia de 100 metros cuando se le aplican los frenos.

(a) ¿Cuál es la fuerza constante producida por los fre. Nos? (b) Con esta fuerza, ¿cuál seria la distancia de frena do si la velocidad hubiera sido de 36 km/h?
Physics
1 answer:
Flura [38]3 years ago
6 0

Answer:

a)    F = 2000 N , B)   x = 25 m

Explanation:

a) To solve this exercise we can use the relationship between work and kinetic energy

          W = ΔK

The job is

           W = -F x

the negative sign is because the force of the brakes is contrary to the movement

as the car stops its final kinetic energy is zero

           K = ½ m v²

let's substitute

         - F x = 0 - ½ m v²

            F = ½ m v² / x

Let's reduce the magnitudes to the SI system

           v = 72 km / h (1000m / 1km) (1 h / 3600 s) = 20 m / s

let's calculate

          F = ½ 1000 20²/100

          F = 2000 N

b) x = ½ mv2 / F

let's slow down to the SI system

         v = 36 km / h = 10 m / s

let's calculate

             x = ½ 1000 10²/2000

             x = 25 m

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3 years ago
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In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

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Answer:

Explained below

Explanation:

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