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9966 [12]
4 years ago
14

In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu

late the x-component of the electric field. Express your answer in terms of the given quantities.b) Calculate the y-component of the electric field. Express your answer in terms of the given quantities.c) Calculate the z-component of the electric field. Express your answer in terms of the given quantities.d) At which point is the electric field equal to zero? 1-) x=0, y=0, z=0 2-) x=-C/A, y=0, z=-2BC/A^2 3-) x=-C/A, y=-2BC/A^2, z=C/A
Physics
1 answer:
laiz [17]4 years ago
6 0

Answer:

a)  Eₓ = - A y + 2B x , b)  Ey = -Ax –C , c) Ez = 0 , d) The correct answer is 3

Explanation:

The electric field and the electric power are related

                    E = - dV / ds

a) Let's find the electric field on the x axis

                  Eₓ = - dV / dx

                  dV / dx = A y - B 2x

                  Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

                Ey = - dV / dy

                dV / dy = A x + C

                Ey = -Ax –C

c) the electric field on the z axis

              dv / dz = 0

              Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

              0 = = - A y + 2B x

              y = 2B / A x

Axis y

             0 = -Ax –C

              .x = -C / A

We substitute this value in the previous equation

             .y = 2B / A (-C / A)

             .y = 2 B C / A2

The correct answer is 3

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A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going
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Answer:

28.6260196842 m

Explanation:

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s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2

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Here, distance traveled by the stone and sound is equal

\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0

t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds

The time taken to fall down is 2.4158 seconds

h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m

Height of the buidling is 28.6260196842 m

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