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Lana71 [14]
3 years ago
9

Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the mag

netic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Physics
1 answer:
LekaFEV [45]3 years ago
8 0

Answer:

The wave is traveling in the y axis direction

Explanation:

Because the wave will always travel in a direction 90° to the magnetic and electric components

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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
Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric fiel
Delicious77 [7]

Answer:

q₁ = q₂ = Q = 14.8 pC

Explanation:

Given that

q₁ = q₂ = Q = ?

Distance between charges = r =7.3 cm = 0.073 m

Combined electric field = E₁ + E₂ = E = 50 N/C

Using formula

E=2\frac{kQ}{r^2}

Rearranging for Q

Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}

Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC

4 0
3 years ago
What conversion takes place in a motor?
LenaWriter [7]
C should be the right answer! hopefully this helps!
4 0
3 years ago
A radio wave has a frequency of 8.6 × 108 hz. what is the energy of one photon of this radiation (h = 6.63 × 10–34 j • s)?
My name is Ann [436]

Answer: 57.018\times10^{-26}J

Energy is directly proportional to the frequency.

The energy of a photon is given by:

E=h\nu

Where, E is the energy and \nu is the frequency.

Frequency of the radio wave is given:

\nu=8.6\times10^8 Hz

h=6.63\times10^{-34}J.s

Multiply the above two:

Energy,

E=h\nu=6.63\times10^{-34}J.s\times8.6\times10^8 Hz= 57.018\times10^{-26}J

Hence, the energy of each photon of radio wave having frequency  \nu=8.6\times10^8 Hz is 57.018\times10^{-26}J

5 0
3 years ago
1. An object 20 cm away from a lens produces a focused image on a film 15 cm away, what is the focal length
morpeh [17]

8.6 cm

Explanation:

Step 1:

In this we have to find the focal length of converging lens.

To find focal length we have,

(1/u) + (1/v) = (1/f)

where u = Object distance

           v= Image distance

           f = Focal length

Step 2:

(1/f) = (1/20) + (1/15)

(1/f) =  0.116

f = 1/0.116

f = 8.6 cm

6 0
3 years ago
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