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Wittaler [7]
3 years ago
11

Two point sources are simultaneously creating waves with a wavelength of 6 cm. Point X lies between the two sources. It is 9 cm

from one source and 15 cm from the other. What wave heights do you expect at point X? A. The waves will never get very high or low. B. The waves may get very high, but will never get very low. C. The waves may get very low, but will never get very high. D. The waves will sometimes get very high and very low.
Physics
1 answer:
jok3333 [9.3K]3 years ago
6 0

Answer:

OPTION D (The waves will sometimes get very high and very low) is the answer.

Explanation:

Wavelength = velocity ÷ frequency

As the frequency which measures the number of waves per unit of time is inversely proportional to the wavelength, point X which lies between two sources, and one source is shorter than another, the wave heights at point x will vary as the distances from point X vary too. This means that waves at point X depending on the wave type and source will get very high at times and very low.

                                                   

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A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point
Marysya12 [62]
<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E   ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶  = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E =  2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

<em>Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;</em>

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

3 0
3 years ago
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