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Nataly [62]
3 years ago
14

Suppose that light from a laser with wavelength 633 nm is incident on a thin slit of width 0.500 mm. If the diffracted light pro

jects onto a screen at distance 1.50 m, what is the distance y2 from the center of the diffraction pattern to the dark band with m=2?
Physics
1 answer:
coldgirl [10]3 years ago
3 0

Explanation:

It is given that,

Wavelength, \lambda=633\ nm=633\times 10^{-9}\ m

Slit width, a=0.5\ mm=0.0005\ m

Order, m = 2

If the diffracted light projects onto a screen at distance 1.50 m, L = 1.5 m

For the diffraction of light,

y=\dfrac{m\lambda L}{a}

y=\dfrac{2\times 633\times 10^{-9}\times 1.5}{0.0005}

y = 0.0037 m

So, the distance from the center of the diffraction pattern to the dark band is 0.0037 meters. Hence, this is the required solution.

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A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy
m_a_m_a [10]

Answer:

(A) Spring constant will be 126.58 N/m

(B) Amplitude will be equal to 0.177 m

Explanation:

We have given mass of the block m = 200 gram = 0.2 kg

Time period T = 0.250 sec

Total energy is given TE = 2 J

(A) For mass spring system time period is equal to T=2\pi \sqrt{\frac{m}{K}}

So 0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}

0.0398=\sqrt{\frac{0.2}{K}}

Now squaring both side

0.00158=\frac{0.2}{K}

K = 126.58 N/m

So the spring constant of the spring will be 126.58 N/m

(B) Total energy is equal to TE=\frac{1}{2}KA^2, here K is spring constant and A is amplitude

So 2=\frac{1}{2}\times 126.58\times A^2

A^2=0.0316

A = 0.177 m

So the amplitude of the wave will be equal to 0.177 m

6 0
3 years ago
Newton's First Law of Motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon
Leno4ka [110]

Law of inertia would be your answer.

4 0
3 years ago
Read 2 more answers
How many neutrons does potassium have?
Sloan [31]

Answer:

the answer is 20 neutrons

Explanation:

6 0
2 years ago
Read 2 more answers
A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the
olya-2409 [2.1K]

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

f = frequency of the tone = 329.6 Hz

T = Time period of the sound wave

we know that, Time period and frequency are related as

T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s

v = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m

v = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m

8 0
3 years ago
Technician A says that the evacuation process will remove dirt and debris from the refrigerant system. Technician B says that th
GuDViN [60]

Answer: Technician B is right.

Explanation:

Evacuation process is used in refrigeration systems to remove moisture, air and non-profit condensable gases in order to achieve maximum function of the system.

vacuum pump is used to draw the sealed AC system into a vacuum. Evacuation of a refrigerant system also helps to maintain pressure, this is so as pulling a vacuum on the system is simply removing matter (mostly air and nitrogen) from inside the system so that the pressure inside drops below atmospheric pressure.

3 0
3 years ago
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