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Mrrafil [7]
3 years ago
10

Sabrina decided to start selling lemonade on her street. The other kids in the neighborhood noticed that Sabrina was making a lo

t of money selling lemonade. These kids decided to open their own lemonade stand. When they opened their own lemonade stand, the equilibrium price ________ and the equilibrium quantity ________. Group of answer choices a. decreased; decreased b. stayed the same; stayed the same c. increased; decreased d. increased; increased e. decreased; increased

Physics
1 answer:
suter [353]3 years ago
6 0

Answer:

e. Decreased; Increased

Explanation:

The market equilibrium refers to the condition in which the price and the quantity are in equilibrium if the conditions of supply (what the producers offer) and demand (what the consumers buy) don't change.

In this case we have that as the other kids decide to sell more lemonade the amount of lemonade is going to increase (equilibrium quantity) and because of this the price is going to decrease (equilibrium price).

Then when the other kids open theirown lemonade stand, the equilibrium price <u>descreased</u> and the equilibrium quantity <u>increased.</u>

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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
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Answer:

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

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

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            |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}

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            |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

               

                         

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