All categories

4 years ago

Question 12 when an electrical pulse on one wire is accidentally detected on another nearby wire, this is known as _____.

2 answers:

8 0

An electric current has an associated magnetic field when the flow occurs in. This is also known as magnetic effect of current which has been<span> observed and quantified by </span>Ampere's Law<span>.

</span>Meanwhile, Current is produced in a conductor when it is moved through a magnetic field because the magnetic lines of force are applying a force on the free electrons in the conductor and causing them to move. This process of generating current in a conductor by placing the conductor in a changing magnetic field is called induction<span>.
</span>
Hence, when a electric wire is placed nearby another wire carrying current, due to magnetic effect of current in the wire carrying electric current, current is induced in the wire. Hence the electric pulse and it is known as induction.

stellarik [79]4 years ago

3 0

Answer:

when an electrical pulse on one wire is accidentally detected on another nearby wire, this is known as Magnetic effect of current

Explanation:

magnetic effect of electric current is known as electromagnetic effect. we can observe that when a compass is brought near a current carrying conductor the needle of compass gets deflected because of flow of electricity. This phenomenon shows that electric current produces a magnetic effect.

You might be interested in

The resolving power of a microscope is proportional to the wavelength used. A resolution of 1.0 10-11 m (0.010 nm) would be requ

Anon25 [30]

Answer:

K = 13448.64eV

Explanation:

(a) In order to calculate the kinetic energy of the electrons, to "see" the atom, you take into account that the wavelength of the electrons must be of the order of the resolution required (0.010nm).

Then, you first calculate, by using the Broglies' relation, the momentum of the electron associated to a wavelength of 0.010nm:

$p=\frac{h}{\lambda}$ (1)

p: momentum of the electron

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 0.010nm

You replace the values of the parameters in the equation (1):

$p=\frac{6.262*10^{-34}Js}{0.010*10^{-9}m}=6.262*10^{-23}kg\frac{m}{s}$

With this values of the momentum of the electron you can calculate the kinetic energy of the electron by using the following formula:

$K=\frac{p^2}{2m}$ (2)

m: mass of the electron = 9.1*10^-31 kg

$K=\frac{(6.262*10^{-23}kgm/s)^2}{2(9.1*10^{-31}kg)}=2.15*10^{-15 }J$

In electron volts you obtain:

$2.15*10^{-15}J*\frac{6.242*10^{18}eV}{1J}=13448.64eV$

The kinetic energy required for the electrons must be, at least, of 13448.64 eV

5 0

3 years ago

A lava flow is an example of what igneous rock

nydimaria [60]

The answer is extrusive.

7 0

4 years ago

Read 2 more answers

Find the horizontal and vertical components of a projectile that is fired with a velocity of 50m/s at 30 degrees relative to the

VashaNatasha [74]

Find this answer on safari and if you can’t reply to me and i’ll help

5 0

3 years ago

In a Rutherford scattering experiment, an alpha particle is accelerated toward a stationary silver nucleus. If the closest appro

lys-0071 [83]

Answer:

Initial Kinetic energy of alpha particle is 9.45x10⁻¹³ J .

Explanation:

The distance at which the initial kinetic energy of the particle is equal to the potential energy is known as closest distance. As it is Rutherford scattering, so it is a coulomb potential energy.

Let K be the initial kinetic energy of alpha particle and r be the closest approach distance. So,

Initial Kinetic Energy = Coulomb Potential Energy

K = $\frac{k\times2e\times{Ze}}{r }$

Here, k is constant, e is charge of electron and Z is the atomic number of silver.

Put 9x10⁹ N m²/C² for k, 1.6x10⁻¹⁹ C for e, 47 for Z and 22.9x10⁻¹⁵ m for r in the above equation.

K = $\frac{9\times10^{9}\times{1.6}\times10^{-19}\times{1.6}\times10^{-19}\times2\times47}{22.9\times10^{-15} }$

K = 9.45x10⁻¹³ J

3 0

4 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago