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Whitepunk [10]
2 years ago
4

A diffraction grating with 355 lines/mm is 1.2 m in front of a screen. What is the wavelength of light whose first-order maxima

will be 16.4 cm from the central maximum on the screen?
Physics
1 answer:
kvv77 [185]2 years ago
6 0

Answer:

The wavelength of light is 381 nm.

Explanation:

Given that,

Distance of screen D= 1.2 m

Diffraction grating = 355 lines/mm

Order m= first

Distance from the central maximum on the screen y= 16.5 cm

We need to calculate the width of slits

d=\dfrac{1}{N}

d=\dfrac{1}{355\times10^{-3}}

d=2.816\times10^{-6}\ m

We need to calculate the angle

Using formula of distance

y=D\tan\theta

\thata=\tan^{-1}(\dfrac{y}{D})

Put the value into the formula

\theta= \tan^{-1}(\dfrac{0.164}{1.2})

\theta=7.78^{\circ}

We need to calculate the wave length

Using formula of wavelength

d\sin\theta=m\lambda

\lambda=\dfrac{d\sin\theta}{m}

Put the value into the formula

\lambda=\dfrac{2.816\times10^{-6}}{1}\sin7.78

\lambda=381\ nm

Hence, The wavelength of light is 381 nm.

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Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

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= 90 /  .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B)  Rest length of the rocket = length of platform = 90 m

C ) length of platform  as viewed by moving observer =

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for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }

8.75 x 10⁻⁸ s .

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2 years ago
An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc
Afina-wow [57]
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

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5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds

Now, we may use this time to determine the horizontal distance covered by the parcel by using:
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The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

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Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.
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The hoop is attached.

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We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ

Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
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Now, solve for θ:
θ = tan⁻¹(2·μ)
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2 years ago
A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acc
MAVERICK [17]
Change in velocity = d(v)
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Change in time = d(t)
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sleet_krkn [62]
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2 years ago
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