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ziro4ka [17]
3 years ago
9

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3 what is the density of gold

Physics
2 answers:
Contact [7]3 years ago
7 0

Answer:

Density, d=19.3\ g/cm^3

Explanation:

Mass of the sample of gold, m = 38.6 grams

Volume of gold sample, V=2\ cm^3

The density of gold sample is given by mass per unit volume. Mathematically, it is given by :

d=\dfrac{m}{V}

d=\dfrac{38.6\ g}{2\ cm^3}

d=19.3\ g/cm^3

So, the density of the sample of gold is 19.3\ g/cm^3. Hence, this is the required solution.

Rudik [331]3 years ago
3 0
Density = Mass ÷ Volume

Your mass = 38.6 grams
Your volume = 2 cm3

38.6 ÷ 2 = 19.3 grams per cm3
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Complete Question

The diagram for this question is shown on the first uploaded image

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The minimum velocity of A is  v_A= 4m/s

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From the question we are told that

    The length of the string is  L = 1m

     The initial speed of block A is u_A

     The final speed of block A is  v_A = \frac{1}{2}u_A

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       m_A u_A  = m_Bv_B + m_A \frac{u_A}{2}

      m_A u_A  - m_A \frac{u_A}{2} = m_Bv_B

          m_A \frac{u_A}{2} = m_Bv_B

  making v_B the subject of the formula

             v_B =m_A \frac{u_A}{2 m_B}

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               v_B =\frac{7 u_A}{4}  

This v__B is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of v__B' at  the top of the vertical circle  

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                   a= \frac{v^2_{B}'}{L}

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         T + mg = m \frac{v^2_{B}'}{L}

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

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                 \frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L

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            v_A = \sqrt{\frac{80mgL}{49m} }

substituting values

          v_A = \sqrt{\frac{80* 9.8 *1}{49} }

              v_A= 4m/s

     

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