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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

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A racecar on a straight track, starting from rest*, steps on the

kirill115 [55]

I think the answer is C

6 0

3 years ago

You drop your frozen rock from a green bridge. The frozen rock starts from rest (initial velocity = 0ms). The rock takes 4.3s to

valentinak56 [21]

Answer:

The velocity of the frozen rock at $t = 1.5\,s$ is -14.711 meters per second.

Explanation:

The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity ( $v$ ), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:

$v = v_{o} + g\cdot t$ (Eq. 1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$g$ - Gravity acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 0\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $1.5\,s$ , then final velocity is:

$v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (1.5\,s)$

$v = -14.711\,\frac{m}{s}$

The velocity of the frozen rock at $t = 1.5\,s$ is -14.711 meters per second.

5 0

2 years ago

A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

Ilya [14]

The frequency of the wave is $1\cdot 10^9 Hz$

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

$c=f \lambda$

where

c is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the radio wave in this problem,

$\lambda = 0.3 m$

$c=300,000,000 m/s = 3\cdot 10^8 m/s$

Therefore, the frequency is:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz$

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0

2 years ago

We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.2

monitta

Answer:

The number of atoms is $N = 4.37*10^{22} \ atoms$

Explanation:

From the question we are told that

The mass of coin is $m_n = 4.2g$

Number of atom in one mole = $n =6.02*10^{23} \ atoms$

Molar mass of nickel $M = 57.8g$

Now the relation to obtain the number of atom in the nickel coin is

$N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel } * No\ of\atoms \ in \ \ one\ mole\ of\ nickel$

$= \frac{4.2}{57.8}* 6.02*10^{23}$

$=4.37 *10^{22} atoms$

8 0

3 years ago

THIS MARCIN

nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

−

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

′

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

′

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

−

(−25)

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

(−25)

−

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

250

25−10

250

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

250

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

′

=−

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0

2 years ago