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ser-zykov [4K]
3 years ago
8

In young's double silt experiment if the distance between the silts is 0.5 and the distance between the silts and screen is 2 ti

mes.Then what will be the width of bands

Physics
1 answer:
maw [93]3 years ago
5 0

Answer: The width of bands will be 2λ

Explanation: Please see the attachments below

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Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the sou
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Answer:

<em>d. unchanged.</em>

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

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Which of the following is the most reasonable weight in units of newtons for an average adult?
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Answer: The reason weight for average adult is 530N

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Work is done on an object when an applied force causes the object to move in the same direction as the force.
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Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame v_{m} 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = \frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}

X = \frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}

X = 8.07 c

Now,

Y = y = - 2

Z = z = 3

Now,

t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}

t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s

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3 years ago
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