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lilavasa [31]
4 years ago
5

Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at

point P, what should be the path difference between the two waves?
The path difference between the two waves should be one and one-quarter of a wavelengths.
The path difference between the two waves should be two wavelengths.
The path difference between the two waves should be one-half of a wavelength.
The path difference between the two waves should be one wavelength.
The path difference between the two waves should be one-quarter of a wavelength.
The path difference between the two waves should be four wavelengths
Physics
1 answer:
amid [387]4 years ago
3 0
"The path difference between the two waves should be one-quarter of a wavelength" is the statement among the choices given in the question that describes the <span>path difference between the two waves. The correct option among all the options that are given in the question is the fifth statement or the penultimate statement.</span>
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What are the factors that affect the resistance of a wire?
777dan777 [17]

1) Length of the wire.

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3 years ago
About how many centimeters will make an inch?<br> 02<br> O 10<br> 100<br> 200
Tresset [83]
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7 0
3 years ago
Read 2 more answers
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s
riadik2000 [5.3K]

Answer:

Part a)

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

\tau = I\alpha

here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

now we will have inertia of two masses given as

I = (m_1 + m_2)(\frac{L}{2})^2

now we have

I = (m_1 + m_2)\frac{L^2}{4}

now the angular acceleration is given as

\alpha = \frac{\tau}{I}

so we have

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

so we will have

I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

now the acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

7 0
4 years ago
A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co
lidiya [134]

The position vector can be transcribed as:

A<span> = 6 i + y j                           </span>

i <span>points in the x-direction and j points in the y-direction.</span>

The magnitude of the vector is its dot product with itself:

<span>|A|2 = A·A</span>

<span>102  = (6 i + y j)•(6 i+ y j)            Note that i•j = 0, and  i•i  = j•j = 1 </span>

<span>100  = 36 + y2       </span>

<span>64    = y2</span>

<span>get the square root of 64 = 8</span>

<span>The vertical component of the vector is 8 cm.</span>

3 0
3 years ago
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