What is the amplitude of photo as they passed a black hole

A point charge is at the origin. With this point charge as the source point, what is the unit vector r^ in the direction of (a)

KiRa [710]

Answer:

a. $\hat{r} =- \hat{j}$
b. $\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$
c. $\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}$

Explanation:

Using Coulomb's Law we know that the electric field E at point $\vec{r}$ is:

$\vec{E(\vec{r})} = k_e \frac{q}{d^2} \frac{\vec{r}-\vec{r'}}{d}$

where $k_e$ is the Coulomb's Constant, q is the source charge, d is the distance between point and position of the source point charge, and $\vec{r}'$ is the position of the source point charge.

Taking all this in consideration, the unit vector clearly is:

$\hat{r} =\frac{\vec{r}-\vec{r'}}{d}$

For our problem, $\vec{r'} = (0,0)$ , as the charge is located at the origin.

So

$\hat{r} =\frac{\vec{r}}{d}$

and d will be the magnitude of $\vec{r}$

Now, we can take the values for each point.

$\vec{r}= (0,-1.35 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(0 \ m)^2 + (-1.35 \ m )^2}$

$|\vec{r}| =1.35 \ m$

So, the unit vector is:

$\hat{r} =\frac{(0,-1.35 \ m)}{1.35 \ m}$

$\hat{r} =(0,-1,0)$

$\hat{r} =- \hat{j}$

$\vec{r}= (12 \ cm,12 \ cm)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(12 \ cm)^2 + (12 \ cm )^2}$

$|\vec{r}| = \sqrt{2} \ 12 \ cm$

So, the unit vector is:

$\hat{r} =\frac{(12 \ cm,12 \ cm)}{\sqrt{2} \ 12 \ cm}$

$\hat{r} =(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$

$\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$

$\vec{r}= (-1.10 \ m, 2.60 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(-1.10 \ m)^2 + (2.60 \ m)^2}$

$|\vec{r}| = 2.8415 \ m$

So, the unit vector is:

$\hat{r} =\frac{ (-1.10 \ m, 2.60 \ m)}{2.8415 \ m}$

$\hat{r} =(-0.3871 ,0.91501)$

$\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}$

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We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
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Gravity 8m/s2 will be at 658 kms from surface of earth.
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What is the amplitude of photo as they passed a black hole​

What is the amplitude of photo as they passed a black hole