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1 year ago

Why is the contribution of the wavelets lying on the back of secondary wave front zero?

Physics

2 answers:

Tresset [83]1 year ago

3 0

Answer:

The contribution of the wavelets lying on the back of the wave front is zero because of something known as the Obliquity Factor. It is assumed that the amplitude of the secondary wavelets is not independent of the direction of propagation, Sources: byju's.com

Anna [14]1 year ago

3 0

According to the Voigt and Kirchhoff's law for wavelets, the contribution of a wavelet in any direction making an angle x with the normal to the wavelet is proportional to {1+cos(x)/2)}

For wavelet lying on the back of secondary wavefront,

x = 180°

→ So, the proportional factor:

→ {1+cos(180°)/2}

→ (1-1/2)

→ (0/2)

→ 0

That's why the contribution of the wavelets lying on the back of secondary wave front is zero

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Pavlova-9 [17]

Answer:

12m

Explanation:

To obtain the answer to the question given, we must observe the characteristics of image formed by a plane mirror.

The image formed by a plane mirror have the following characteristics:

1. Laterally inverted.

2. Same distance as the object from the mirror.

3. Same height as the object.

4. Virtual.

With the above information, we can calculate the distance between the boy and his image as follow:

Initially:

Object distance (u) = 4m

Image distance (v) = 4m

The boy moved 2m away, therefore:

Object distance (u) = 2 + 4 = 6m

Image distanc(v) = 2 + 4 = 6m

The distance between the boy and his image will be the sum of his distance (u) and image distance (v) i.e (u + v)

The distance between the boy and his image = 6 + 6 = 12m

Therefore, the distance between the boy and his image is 12m.

6 0

2 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$