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dexar [7]
3 years ago
5

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

balanced? The answer is 13.4 In the mobile what is the value for m3 to the nearest hundredth of a kilogram?

Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

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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3 years ago
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

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Answer:

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Explanation:

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