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Law Incorporation [45]

3 years ago

10

Why does the pencil appear broken in the cup of water?

2 answers:

yulyashka [42]3 years ago

7 0

Electromagnetic/Light waves travel at different speeds through different media.This results in refraction which gives the appearance of broken pencil to our eyes

Sedaia [141]3 years ago

6 0

Answer: Electromagnetic/Light waves travel at different speeds through different media.

Explanation: The reason for the pencil to appear broken is because because of the process of refraction. Light travels at different speeds in different medium because of the varying densities, here the two mediums being air and water.

The difference in speed results in an apparent shift of the position of part of pencil under water and thus the part appears broken.

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Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

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3 0

3 years ago

A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minute

Rudik [331]

Answer:

W = 2352 J

Explanation:

Given that:

mass of the bucket, M = 10 kg

velocity of pulling the bucket, v = 3 $m.min^{-1}$

height of the platform, h = 30 m

time taken, t = 10 min

rate of loss of water-mass, m = $0.4 kg.min^{-1}$

Here, according to the given situation the bucket moves at the rate,

$v=3 m.min^{-1}$

The mass varies with the time as,

$M=(10-0.4t) kg$

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x = 3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

<u>For work calculation:</u>

$W=\int_{0}^{10} [(10-0.4t).g\times 3] dt$

$W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}$

$W= 2352 J$

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Explanation:

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