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sesenic [268]
4 years ago
12

An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.7 × 10-16 N acts

on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 22 mm horizontally.
Physics
1 answer:
anzhelika [568]4 years ago
3 0

Explanation:

GIven data :

Speed V = 1.1×10⁷ m/s

Force F = 3.7×10⁻¹⁶N

Mass of electron = m = 9.11×10⁻³¹Kg

Horizontal Distance X = 2.2×10⁻²m

Vertical distance Y = ?

Solution:

As we know that,

F =ma

where a is the acceleration of electron.

(3.7×10⁻¹⁶) = (9.11×10⁻³¹) a

a = 4.06×10¹⁴m/s² upward.

∵opposing weight of electron is negligible.

now we find how long it takes the electron to move 22mm horizontally.

X =V×T

T = X/V

T =2.2×10⁻²/1.1×10⁷

  = 2×10⁻⁹s

now we can find out the vertical distance Y.

Y = 0.5aT²

  = (0.5)(4.06×10¹⁴)(2×10⁻⁹)²

  = 8.12×10⁻⁴m

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Will reflect ........
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4 years ago
Mirage is due to...
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<span>Mirage is due to (A). unequal heating of different parts of the atmosphere,
because that produces unequal density of different parts of the atmosphere,
which bends the light.
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3 0
4 years ago
You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
wlad13 [49]

Answer:

  • between locations that are 14 cm outboard of the chair edges
  • the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

  (50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

  67(1) +54(x) = 50(1.5)

  x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

  67(1.5) = 54(x) +50(1)

  x = 50.5/54 ≈ 0.935 . . . . meters

<u>Scenario 1</u>

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

<u>Scenario 2</u>

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

__

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

<em>Comment on the question</em>

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

3 0
4 years ago
PLEASE HELP A student climbs to the top of a press box where the cameras are. He wonders how many meters he is off of the ground
Feliz [49]

The height of the rail on top of the press box where the ball was dropped from is 11.025 m.

The given parameters:

  • Time of motion of the ball, t = 1.5 s
  • Let the height of the rail = h

<h3>Maximum height of fall;</h3>
  • The maximum height through which the ball was dropped from is calculated by applying second equation of motion;

h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = \frac{1}{2} (9.8) (1.5)^2\\\\h = 11.025 \ m

Thus, the height of the rail on top of the press box where the ball was dropped from is 11.025 m.

Learn more about height of projectiles here: brainly.com/question/10008919

3 0
2 years ago
A 4.0 Ohms resistor, an 8.0 Ohms resistor, and a 12.0 Ohms resistor are connected in series with a 24.0 V battery. Calculate the
Flura [38]

Answer:

RESISTANCE=(4+8+12)Ohms

=24 ohms

V=IR

24= I ×24

I= 24/24

I= 1 A

CURRENT IS 1 A

8 0
3 years ago
Read 2 more answers
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