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Troyanec [42]
3 years ago
10

How can electricty be used to create a magnetic field

Physics
2 answers:
Naya [18.7K]3 years ago
5 0
A magnetic field pulls and pushes electrons in some objects near them to make them move metals like copper have electrons that are moved easily and can be readily moved from their orbits if a magnet is moved quickly through coil of copper wire electrons move and electricity is made u can use this but maybe u will want to shorten it down
Strike441 [17]3 years ago
4 0
When a charged particle such as an electron, proton or ion is in motion, magnetic lines of force rotate around the particle. Since electrical current moving through a wire consists of electrons in motion, there is a magnetic field around the wire.      i hope u have smileing day

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PLEASE HELP:
german

Answer:

3.39 mins.

10m/s multiplied by 6 equals to 60 seconds or 1 minute

7 0
2 years ago
An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
10 turns of wire are closely wound around a pencil as shown in the figure. when measured using a scale as shown, the length of t
Mila [183]

Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = \frac{length of total turns}{number of turns}

                           = \frac{25}{10}

                           = 2.5 mm

Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

10 mm = 1 cm

So that,

2.5 mm = x

x  = \frac{2.5}{10}

   = 0.25 cm

The wire is 0.25 cm thick.

8 0
3 years ago
How do you solving kinematic equations for horizontal projectiles?
daser333 [38]
See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION

7 0
2 years ago
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
3 years ago
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