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ELEN [110]
3 years ago
13

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 screendeflected out of the plane of screendeflected toward top of the screendeflected into plane of screenunaffected in its direction of motion
Physics
1 answer:
muminat3 years ago
8 0

Answer:

deflected toward bottom of the screen

Explanation:

When entering the region with magnetic field, a magnetic force is exerted on the proton. This force is perpendicular to both the direction of the magnetic field and the direction of the velocity of the proton.

The direction of the force can be determined by using the right-hand rule. We have:

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

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

- Thumb: direction of the magnetic force --> toward bottom of the screen

So, the correct answer is

deflected toward bottom of the screen

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A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar
sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

v^2-u^2 = 2gd

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u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

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Solving for h,

v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

y_1 = h - \frac{1}{2}gt^2

While the vertical position of the ball thrown upward is

y_2 = ut - \frac{1}{2}gt^2

The two balls meet when

y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

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So the distance below the top of the cliff is

d=24.0 - 18.0 = 6.0 m

4 0
3 years ago
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