All categories

3 years ago

A -3.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

Physics

1 answer:

Korvikt [17]3 years ago

4 0

Answer:

a) -537 N/C , b) - 327.8 N/C and c) 723.7 N/C

Explanation:

The electric field is a vector magnitude, so we must add them as vectors. The electric field equation is

E = k q / r²

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m²/C², that the magnitude of the load and r the distance between the load and the test load

Let's find the field created by each charge at the point x = 0.200 m

charge 1.

This charge is at the origin, the distance is

x₁ = 0.200 m

q₁ = 3 nC = 3 10-9 C

E1 = k q₁ / x₁²

E1 = 8.99 10⁹ 3 10⁻⁹ / 0.2²

E1 = 674.25 N / C

Charge 2

This load is the point a = 0.800 m, so the distance to the test charge at 0.200 m

x₂ = 0.800 - 0.200

x₂ = 0.600 m

q₂ = -5.5 nC = -5.5 10-9 C

E2 = 8.99 10⁹ 5.5 10⁻⁹ / 0.600²

E2 = 137.35 N / C

We already have the value of each field, add them vectorially, remember that if the charges are of the same sign they repel and if they are of the opposite sign they attract, the field E1 is directed to the left and the field E2 is directed to the right

Et = -E1 + E2

Et = -674.25 + 137.25

Et = -537 N / C

The field is headed to the left

b) we perform the same procedure for another distance value

Charge 1

x = 1.20

E1 = 8.99 10⁹ 3 10⁻⁹ / 1.2²

E1 = 18.73 N / C

Chage 2

x = 0.8 - 1.2 = -0.4 m

E2 = 8.99 10⁹ 5.5 10⁻⁹ / 0.4²

E2 = 309.03 N / C

Total field, E1 is on the left and E2 goes on the left

Et = -E1 -E2

Et = -18.73 - 309.03

Et = - 327.8 N / C

The field is headed to the left

c) point at x = -0.200 m

charge 1

x = -0.200 m

E1 = 8.99 10⁹ 3 -10⁻⁹ / (-0.2)²

E1 = 674.25 N / C

charge 2

x = 0.800 - (-0.2) = 1,000 m

E2 = 8.99 10⁹ 5.5 10⁻⁹ / 1²

E2 = 49.45 N / C

Ei goes to the right and E2 goes to the right

Et = E1 + E2

Et = 674.25 +49.45

Et = 723.7 N / C

The field is headed to the right

You might be interested in

What causes irregular galaxies to be irregular

Mariulka [41]

Irregular galaxies get their odd shapes in many ways. One way irregular galaxies are formed is when galaxies collide or come close to one another, and their gravitational forces interact. Another source of irregular galaxies may be very young galaxies that have not yet reached a symmetrical state.

8 0

3 years ago

Read 2 more answers

On Mars gravity is one-third that on Earth. What would be the mass on Mars of a person who has a mass of 90 kilograms (kg) on Ea

snow_tiger [21]

Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

4 0

3 years ago

A certain runner averages 4.1 m/s over a 10

Ghella [55]

Answer: 40.650406504065 or 40 minutes and 39 seconds.

Explanation:

1 k = 1000m

race = 10000m

runner time = 10000 / 4.1

runner time = 2439.0243902439024 seconds

runner time = 2439.0243902439024/60 = 40.650406504065 or 40 minutes and 39 seconds.

6 0

3 years ago

Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th

dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

Let's write the diffraction equation for a slit

a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

θ = λ / a

In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

θ = 1.22 λ / D

Where D is the diameter of the pupil

Let's apply this equation to our case

θ = 1.22 600 10⁻⁹ / 0.460 10⁻²

θ = 1.591 10⁻² rad

This is the angle separation to solve the two light sources

6 0

3 years ago

1. An object on Earth and the same object on the Moon would have a difference in

Feliz [49]

Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass $m$ of the body by the acceleration of gravity $g$ :

$W=m.g$

Then, since the Earth and the Moon have different values of gravity, t<u>he weight of an object in each place will vary</u>, but its mass will not.

<h2>Answer 2: b. Force changes by 2/9</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

If we double the mass of one object (for example $2m_{1}$ ) and triple the distance between both (for example $3r$ ). The equation (1) will be rewritten as:

$F=G\frac{2m_{1}m_{2}}{(3r)^2}$ (2)

$F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}$ (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:

$W=(F)(d)$

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

The Kinetic Energy is given by:

$K=\frac{1}{2}mV^{2}$ (4)