1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lemur [1.5K]
3 years ago
9

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

The distance d between both lines is 4.10 m. What is the magnitude of the electric field at a point equidistant from the lines, with a distance 2d from both lines?

Physics
1 answer:
Cloud [144]3 years ago
6 0

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

You might be interested in
HELP PLEASE BRAINLIEST!!!
Dmitrij [34]
<span>The potential energy decreases liquid particles that are attracted to one another move closer together to form a solid. Temperature remains constant because kinetic energy remains constant</span>
6 0
3 years ago
Read 2 more answers
A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir
Pachacha [2.7K]

Answer:

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

\vec{F} = q \vec{E}

11\hat{i} = -5 \times 10^{-6} \vec{E}

\vec{E} =-2.2 \times 10^{6}\hat{i}

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

\vec{F} = q \vec{E}

\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Force experienced by the photon in the same field is equal to \vec{F}= -3.52\times 10^{-13}\hat{i}\ N

3 0
3 years ago
A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir
Fantom [35]

Answer:

linear charge density = -9.495 × 10^{-34} C/m

Explanation:

given data

revolutions per second = 1.80 × 10^{6}

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

so

- q × E = m × w² × r     ..................1

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}r} q =  m × w² × r   ............2

and w = \frac{2*\pi}{T}  

w = \frac{d\theta }{dt}

w = 1.80 × 10^{6} × \frac{2*\pi}{1}

w = 11304000 rad/s

so here from equation 2

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}{0.012} 1.80 × 10^{6} =  1.672 × 10^{-27} × 11304000² × 0.0120  

linear charge density = -9.495 × 10^{-34} C/m

8 0
3 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
4 years ago
A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

$\mathrm{u}=\cos (\omega \mathrm{t})

Then:

$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0
2 years ago
Other questions:
  • Constructive interference will cause two sound waves to cancel each other out resulting in silence
    6·1 answer
  • A pronghorn antelope has been observed to run with a top speed of 97 km/h. Suppose an antelope runs 1.5 km with an average speed
    13·1 answer
  • Based on the information in the graph, which of the atoms listed below is the most stable?
    15·2 answers
  • Isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine are known as the ____.
    5·1 answer
  • What is the most common state of matter in the universe
    14·1 answer
  • Mark all the hadrons.a) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino
    6·1 answer
  • Which is easier for a beginner? (This is for archery)<br><br> O long bow<br><br> O recurve bow
    5·2 answers
  • In the attachment there is a density column where there is colour
    10·2 answers
  • A wave has a frequency of 775 Hz. What is its period?
    13·1 answer
  • Why is it important to have a closed circuit?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!