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4 years ago

15

Calculate the frequency of each of the following wave lengths of electromagnetic radiation. Part A 488.0 nm (wavelength of argon

laser) Express your answer using four significant figures. ν1 ν 1 = nothing s−1 Request Answer Part B 503.0 nm (wavelength of maximum solar radiation) Express your answer using four significant figures.

1 answer:

Drupady [299]4 years ago

7 0

Answer:

A. $f=6.1475*10^{14}Hz$

B. $f=5.9642*10^{14}Hz$

Explanation:

The frequency has an inversely proportional relationship with the concept of wavelength, the greater the wavelength, the lower the frequency. For electromagnetic waves, the frequency is equal to the speed of light, divided by the wavelength.

$f=\frac{c}{\lambda}$

A.

$f=\frac{3*10^8\frac{m}{s}}{488*10^{-9}m}\\\\f=6.1475*10^{14}Hz$

B.

$f=\frac{3*10^8\frac{m}{s}}{503*10^{-9}m}\\\\f=5.9642*10^{14}Hz$

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Doss [256]

Answer:

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Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, $\overrightarrow{v_1}=-120\ \vec{i}$

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Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

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Plug in the given values and solve for the magnitude of the resultant.This gives,

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Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

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fredd [130]

Energy to lift something =

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This simple formula only works if you use the right units.

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For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

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So we have ...

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3 years ago

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

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