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andreev551 [17]
3 years ago
7

The length and width of a rectangular room are measured to be 3.92 ± 0.0035 m and 3.15 ± 0.0055 m. In this problem you can appro

ximate the error on a product or quotient of quantities by the sum of the percent error on each quantity.A.) Calculate the floor area of the room in square meters B.) Calculate the uncertainty in this measurement in square meters.
Physics
1 answer:
Pavel [41]3 years ago
7 0

Answer:

A)A=12.2480\ m^2

B)12.2480\pm 0.1029\ m^2

Explanation:

<u>Given:</u>

Length of the room l= 3.92 ± 0.0035

Width of the room w= 3.15 ± 0.0055

A) Let A be the area of the room

A=l\times w\\A=3.92\times3.15\\A=12.2480\ \rm m^2

B)We will calculate uncertainty in each dimension

%uncertainty in length=\dfrac{0.0035}{3.92}\times 100=0.0892\ %

%uncertainty in width =\dfrac{0.0055}{3.15}\times 100=0.0174%

The uncertainty in area will be sum of uncertainty in length and width

%uncertainty in Area=  %uncertainty in length + %uncertainty in width

%uncertainty in Area=0.0892\ % + 0.0174\ %

%uncertainty in Area=0.0106

Uncertainty in Area=0.0106\times 12.2480=0.1029\ \rm m^2

There Area is12.2480 ± 0.1029\ \rm m^2

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Answer:

For destructive interference phase difference is

(2n+1)\pi where n∈ Whole numbers

Explanation:

For sinusoidal wave the interference affects the resultant intensity of the waves.

In the given example we have two waves interfering at a phase difference of \frac{\pi}{4} would lead to a constructive interference giving maximum amplitude at at the RMS value of the amplitude in resultant.

Also the effect is same as having a phase difference of  ( \frac{\pi}{4} + 2\pi) because after each 2π the waves repeat itself.

<em>In case of destructive interference the waves will be out of phase i.e. the amplitude vectors will be equally opposite in the direction at the same place on the same time as shown in figure.</em>

They have a phase difference of \pi or which is same as (2\pi+\pi)

Generalizing to:

a phase difference of (2n+1)\pi where n∈ {W}

{W}= set of whole numbers.

3 0
3 years ago
How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
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Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{9}{40}

f=4.44\ cm

When , R₁= -4, R₂ = 8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{3}{40}

f=-13.33\ cm

When , R₁= 4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{3}{40}

f=13.33\ cm

When , R₁= -4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{9}{40}

f=-4.44\ cm

Hence, The lenses with different focal length are four.

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