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murzikaleks [220]
3 years ago
15

Un movil avanza a 20 m/s y recorre una distancia de 800 km. Determinar el tiempo en horas que utiliza

Physics
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

t = 11.1 hours

Explanation:

The question says that, "A mobile advances at 20 m / s and travels a distance of 800 km. Determine the time in hours you use".

Given that,

Speed of a mobile, v = 20 m/s = 72 km/h

Distance, d = 800 km

We know that,

Speed = distance/time

So,

t=\dfrac{d}{v}\\\\t=\dfrac{800}{72}\\\\t=11.1\ h

So, it will take 11.1 hours.

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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 }

                m_3 = 0.59 kg

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