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MaRussiya [10]
3 years ago
7

PLEASE ANSWER! WILL MARK BRAINLIEST!

Physics
1 answer:
Angelina_Jolie [31]3 years ago
8 0

The first one would be thermal energy

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Plasma

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Read 2 more answers
A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive
Ivahew [28]

Answer:

a

The  x- and y-components of the total force exerted is

           F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

b

 The magnitude of the force is  

            |F_{31 +32}| = 10.25 *10^{-5} N

   The direction of the force is  

         \theta =327.43 ^o   Clockwise from x-axis

Explanation:

From the question we are told that

    The magnitude of the first charge is q_1 = +5.00nC = 5.00*10^{-9}C

      The magnitude of the second charge is q_2 = -2.00nC = -2.00*10^{-9}C

        The position of the second charge  from the first one is  d_{12} = 4.00i \  cm = \frac{4.00i}{100} = 4.00i *10^{-2} m

        The  magnitude of the third charge is q_3 = +6.00nC = 6.00*10^{-9}C

       The position of the third charge from the first one is  \= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} =  (4i + 3j) *10^{-2}m

                |d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m

                |d_{31}| =5 *10^{-2} m

        The position of the third charge from the second  one is

                \= d_{32} = 3j cm = 3j *10^{-2}m

               |d_{32}| =(\sqrt{ 3^2}) *10^{-2} m

               |d_{32}| =3 *10^{-2} m

The force acting on the third charge due to the first and second charge is mathematically represented as

           F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}

 Substituting values

          F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}  \\ \ +  \ \ \ \ \ \ \ \ \   \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}

            F_{31 +32} = 2.16 *10^{-5} (4i + 3j)  - 12*10^{-5} j

            F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

The magnitude of     F_{31 +32}  is mathematically evaluated as

            |F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}

             |F_{31 +32}| = 10.25 *10^{-5} N

The direction is obtained as

            tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}

              \theta = tan ^{-1} [-0.63889]

             \theta = - 32.57 ^o

             \theta = 360 - 32.57

            \theta =327.43 ^o

               

                         

5 0
3 years ago
A 55-kg woman is wearing high heels. If each heel has a circular cross-section 6.0 mm in diameter and she puts all her weight on
bogdanovich [222]

Answer:

The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

Explanation:

Given;

mass of the woman, m = 55 kg

diameter of the circular heel, d = 6.0 mm

radius of the heel, r = 3.0 mm = 0.003 m

Cross-sectional area of the heel is given by;

A = πr²

A = π(0.003)²

A = 2.8278 x 10⁻⁵ m²

The weight of the woman is given by;

W = mg

W = 55 x 9.8

W = 539 N

The pressure exerted by the woman on the floor is given by;

P = F / A

P = W / A

P = 539 / (2.8278 x 10⁻⁵ )

P = 1.9061 x 10⁷ N/m²

Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

5 0
3 years ago
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