Ask question

Ask question

All categories

3 years ago

14

Multiple-Choice Homework Problem 3.1 [3 pt(s) ] A negatively charged particle is fixed at the origin. What is the direction of t

he electric field at a point along the positive x axis

1 answer:

faltersainse [42]3 years ago

5 0

Answer:

Towards the origin on the positive x-axis.

Explanation:

For a negatively charged particle the direction electric field lines is always into the charge. So, when a negative charge is placed on the origin then the direction of the electric field lines will be 180° to the positive x-axis, i.e. the field lines will be travelling from infinity towards the origin.
Contrary to this the positive charges have the direction of electric field lines emerging out from the charge.

You might be interested in

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

4 years ago

How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n

grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

$dSin\theta = m\lambda$

Where

m= Any integer which represent the number of repetition of spectrum

$\lambda$ = Wavelength

d = Distance between the slits.

$\theta$ = Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

$d = \frac{m\lambda}{sin\theta }$

$d = \frac{2*536*10^{-9}}{sin(24)}$

$d = 2.63*10^{-6}m$

Therefore the number of lines per millimeter would be given as

$\frac{1}{d} = \frac{1}{2.63*10^{-6} }$

$\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})$

$\frac{1}{d} = 379.4 lines/mm$

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0

3 years ago

Please help on this one?

bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0

3 years ago

A protein molecule in an electrophoresis gel has a negative charge. The exact charge depends on the pH of the solution, but 30 e

Reika [66]

Answer:

7.401 * 10^(-15) N

Explanation:

30 electrons will have a charge:

30 * -1.6022 * 10^(-19) C

= - 4.806 * 10^(-18) C

The relationship between electric field and electric force is:

E = F/q

This means that force, F, is

|F| = |E|*|q|

|F| = |1540| * |-4.806 * 10^(-18)|

|F| = |-7401.24 * 10^(-18)|

|F| = 7.401 * 10^(-15) N

7 0

4 years ago

How do I calculate speed, velocity, and acceleration? I need the formulas too

vodomira [7]

Speed is the secular unit, which means it only measures "how fast it goes."
$speed= \frac{distance}{time}$
Velocity is a victory unit, which means it measures "both how fast it goes AND the direction."
$velocity= displacement/time$
Acceleration is the change of velocity over time.
$acceleration = \frac{change in velocity}{time}$

5 0

3 years ago