26 POINTS!! will give brailest! can anyone help me? if you can't well, Merry Christmas!

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in

bezimeni [28]

1) Potential difference: 1 V

2) $V_b-V_a = -1 V$

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

$\Delta U=q\Delta V$

where

q is the charge's magnitude

$\Delta V$ is the potential difference between the initial and final position

In this problem, we have:

$q=4.80\cdot 10^{-19}C$ is the magnitude of the charge

$\Delta U = 4.80\cdot 10^{-19}J$ is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

$\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V$

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

$V_b-V_a = - 1V$

8 0

2 years ago

Which of the following is not an example of centripetal acceleration?

Amanda [17]

An apple falling to the ground is not an example of centripetal acceleration.

5 0

3 years ago

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:
</span>
$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span> $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$