Which graph represents the linear equation y = 1?

Timed math quiz 2 <br> Please help me

Alex73 [517]

Answer:

5

Step-by-step explanation:

everything is multiplied by 5

3 0

3 years ago

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An electron at rest at P = (5,3,7) moves along a path ending at Q = (1, 1, 1) under the influence of the electric field (in newt

EastWind [94]

Answer:

a) $\bf f(x,y,z)=200ln(x^2+z^2)$

b) $\bf 3.9819*10^{16}\;m/seg$

Step-by-step explanation:

The electric field can be written as

$\bf F(x,y,z)=(\displaystyle\frac{400x}{x^2+z^2},0,\displaystyle\frac{400z}{x^2+z^2})$

a)Find a potential function for F

A potential function for F would be a function f such that

$\bf \nabla f = F$

That is, the gradient of f equals F.

If such a function exists then

$\bf \displaystyle\frac{\partial f}{\partial x}=\displaystyle\frac{400x}{x^2+z^2}\\\\\displaystyle\frac{\partial f}{\partial y}=0\\\\\displaystyle\frac{\partial f}{\partial z}=\displaystyle\frac{400z}{x^2+z^2}$

Integrating the first equation with respect to x

$\bf f(x,y,z)=\displaystyle\int\displaystyle\frac{\partial f}{\partial x}dx+h(y,z)$

where h(y,z) is a function that does not depend on x. We have then

$\bf f(x,y,z)=\displaystyle\int\displaystyle\frac{400x}{x^2+z^2}dx+h(y,z)=400\displaystyle\int\displaystyle\frac{x}{x^2+z^2}dx+h(y,z)=\\\\200ln(x^2+z^2)+h(y,z)$

By taking the partial derivatives with respect y and z we can notice that h(z,y)=0.

Therefore, a potential function for F is

$\bf \boxed{f(x,y,z)=200ln(x^2+z^2)}$

b) What is the electron’s speed at the point Q?

Since F is conservative, the work done to move the particle from P to Q does not depend on the path, but only on the potential function f.

Let W be the work done when moving the particle from P to Q

W = f(Q) - f(P) = f(1,1,1) - f(5,3,7)

$\bf W=200ln(1^2+1^2)-200ln(5^2+7^2)=-722.183583$ joules

According to the law of conservation of energy, the work done to move the electron from P to Q equals the change in kinetic energy of the object.

Since the electron is at rest in P, the kinetic energy at P equals 0 and we have

$\bf W=-\displaystyle\frac{m(v_Q)^2}{2}$

where

m = mass of an electron

$\bf v_Q$ = speed at point Q

Replacing in the equation

$\bf -722.183583=-\displaystyle\frac{9.10938356*10^{-31}(v_Q)^2}{2}\Rightarrow\\\\\Rightarrow v_Q=\sqrt{\displaystyle\frac{2*722.183583}{9.10938356*10^{-31}}}=3.9819*10^{16}\;m/seg$