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MrMuchimi
3 years ago
10

Proton traveling across a capacitor A proton enters with a velocity v between the plates of a capacitor as shown in the figure.

What is the direction of the magnetic field needed so that the proton continues its trajectory, undeflected? Explain each step of your reasoning.
Physics
1 answer:
yulyashka [42]3 years ago
3 0

Answer:

Please see the direction of force in the attachment

Explanation:

As the particle enters the capacitor, it experiences electric. In order to move in the same direction as the electron was moving before entering the capacitor, the force due to the electric field of capacitor must be balanced by the magnetic field.

The direction of particle is shown in the attachment.

As we know,

Force = q (v *B)

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The area vector of a square loop of 5 turns of a conductor each with side length of 0.2 m carrying a current of 2 A is antiparal
Anon25 [30]

Answer:

M= 0.4 Am^2

Explanation:

From the question we are told that:

Number of turns  N=5N=5

Conductor each with side length L=0.2m

Current I=2A

Magnetic field  B=50.0T

Generally the equation for the total magnetic moment  M is mathematically given by

M = current * area

M= I * A

M = 2 * (5* 0.2*0.2)  

M = 2 * 0.2

M= 0.4 Am^2

3 0
2 years ago
Why do stars appear so steady when viewed from the surface of moon or<br> by an astronaut in space?
Nutka1998 [239]

Answer:

Explanation:

The light from these little disks is also refracted by Earth's atmosphere, as it travels toward our eyes. That's because, in the direction of any horizon, you're looking through more atmosphere than when you look overhead. If you could see stars and planets from outer space, both would shine steadily.

Answer From GauthMath If you like it then please heart it and comment thanks.

4 0
3 years ago
Will give brainlist.
Lera25 [3.4K]

Answer:

<u>Displacement (km)</u>

Explanation:

The y axis is the vertical axis pointing up and down. This is labeled as the the displacement (km) in the graph.

6 0
3 years ago
When you hit a .27 kg volleyball the contact time is 50 ms and the average force is 125 N. If you serve the volleyball (from res
yulyashka [42]

(A) P(v) = 0.135v

(B) P(h) = 0.234v

<u>Explanation:</u>

Given-

Mass of the ball, m = 0.27kg

Force, F = 125N

angle of projection, θ = 30°

Let v be the velocity of the ball.

A) vertical component of the momentum of the volleyball

We know,

P(vertical) = mvsinθ

P(V) = 0.27 X v X sin 30°

P(V) = 0.27 X v X 0.5

P(V) = 0.135v

B) horizontal component of the momentum of the volleyball

We know,

P(Horizontal) = mvcosθ

P(h) = 0.27 X v X cos 30°

P(h) = 0.27 X v X 0.866

P(h) = 0.234v

8 0
3 years ago
A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h
katen-ka-za [31]

Answer:

v_r=5.89\ m.s^{-1}

Explanation:

Given:

  • mass of rocket, m_r=50\ g
  • time of observation, t=2\ s
  • mass lost by the rocket by expulsion of air, m_a=10\%\ of m_r=5\ g
  • velocity of air, v_a=53\ m.s^{-1}

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

m_a.v_a=(m_r-m_a)\times v_r

5\times 53=(50-5)\times v_r

v_r=5.89\ m.s^{-1}

7 0
3 years ago
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