All categories

3 years ago

The diagram does not represent a real electric field because the field lines, can someone help explain this for me

Physics

1 answer:

Papessa [141]3 years ago

4 0

electric field lines are graphical presentation of electric field intensity

It is the graphical way to represent the electric field variation

If we draw the tangent to electric field line then it will give the direction of net electric field at that point

So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties

here we will always follow these basic properties of field lines

now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined

As we have two directions of tangents at that point

So here the incorrect presentation is the intersection of two field lines which is not possible

You might be interested in

A mason is shot with a constant speed of 7.5 x 10²m/sinto a region, when an electric field produces acceleration on the mason of

nadya68 [22]

Answer:

Distance, d = 0.1 m

It is given that,

Initial velocity of meson,

Finally, the meson is coming to rest v = 0

Acceleration of the meson, (opposite to initial velocity)

Using third equation of motion as :

s is the distance the meson travelled before coming to rest.

So,

s = 0.1 m

The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.

5 0

2 years ago

What is the metric unit of force? kg kg/cm^3 kg/cm^2 N

Feliz [49]

Force is found by multiplying mass (kg) and acceleration (m/s^2), so the metric unit of force is kg*m/s^2 or N (newtons)

4 0

3 years ago

Read 2 more answers

Help ASAP with equation and answer pls. Numbers 2 3 4

scoray [572]

Answer:

2. 200N

3.50kg

4.700N

Explanation:

Weight is another word for the force of gravity

Weight is a force that acts at all times on all objects near Earth.

F=m*g

where g=acceleration due to gravity

2. due to the gravitational fields of the earth , assume gravitational acceleration=10m/s2

F=20*10= 200N

3.same as above

mass=Force/gravitational acceleration

mass=500/10 = 50kg

4.force=mass*gravitational acceleration

force=70*10=700N

8 0

3 years ago

Birds sitting on a single power line don't get shocked. But if they were to place one foot on each of two lines, ___________ wou

Natalka [10]

Current would flow between them and they would receive a terrible shock.

5 0

2 years ago

Read 2 more answers

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago