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nlexa [21]
3 years ago
12

Find the magnitude of the vector v given its initial and terminal points. Round your answer to four decimal places.

Physics
1 answer:
Sliva [168]3 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is |v| = 6.93

Explanation:

From the question we are told that

    The initial point is (x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )

    The  terminal point is  (x_2 , y_2 , z_2) = (-5 , 11, 8 )

Generally the magnitude of the vector is mathematically represented as

     |v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}

=>   |v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}

=>   |v| = 6.93

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A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may
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Answer:

The tension on the string is  T  =  43.302 \ N

Explanation:

From the question we are told that

    The mass of the rock is m_r = 5.00 \ kg =  5000 \ g

       The density of the rock is \rho  =  4300 \ kg/m^3 =  4.3 g/dm^3

       

Generally the volume of the rock is mathematically evaluated as

          V    =  \frac{m_r}{\rho}

substituting values

        V    =  \frac{5000}{4.3}

       V    =  1162.7 \  dm^3

The volume of the rock immersed in water is

      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

     V_w = 581.4 \ dm^3

mass of water been displaced by the this volume is

     m_w  = V_w     According to Archimedes principle

=>   m_w =  581.4 \ g

     m_w =  0.5814 \ kg

The weight of the water displace is  

      W _w =  m_w  * g

      W _w =  0.5814  * 9.8

      W _w = 5.698 \ N

The actual weight of the rock is  

      W_r  =  m_r * g

     W_r  =  5.0 *  9.8

     W_r  =  49.0 \ N

The tension on the string is

       T  = W_r - W_w

substituting values

       T  = 49.0 -  5.698

       T  =  43.302 \ N

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