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Crazy boy [7]
3 years ago
6

A proton traveling to the right enters a region of uniform magnetic field that points into the screen. When the proton enters th

is region, it will be..??
deflected toward bottom of the screen
deflected out of the plane of screen
deflected toward top of the screen
deflected into plane of screen
unaffected in its direction of motion
Physics
1 answer:
Natali5045456 [20]3 years ago
7 0

Answer:

deflected toward top of the screen

Explanation:

We can find the direction of the force acting on the proton by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the proton --> to the right

- Middle finger: direction of the magnetic field --> into the screen

- Thumb: it gives the direction of the force --> therefore, it will be upward

And since the charge is positive (it is a proton), we don't have to revers the direction: so, the force on the proton points upward.

Therefore, the correct answer will be

deflected toward top of the screen

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Answer:

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Explanation:

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A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

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During the daytime, I have mostly line symmetry.

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