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Harrizon [31]
3 years ago
10

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of

2.70 x 10^-3 V/m and is parallel to the x axis. What are the frequency and the magnetic-field amplitude of this wave? In calculating the amplitude of the magnetic field, you divide the electric-field amplitude by the speed of light, which requires you to convert some units. Which of the following is equivalent to 1 T?
a. 1 (N/A.m)
b. 1 (N/c.m)
c. 1 (N.s/C.m)
d. 1 (V.s/m^2)
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

a) 6.9*10^14 Hz

b) 9*10^-12 T

Explanation:

Given that

The wavelength of the wave, λ = 435 nm

Amplitude of the electric field, E(max) = 2.7*10^-3 V/m

a)

The frequency of the wave can be found by using the formula

c = fλ, where c = speed of light

f = c/λ

f = 3*10^8 / 435*10^-9

f = 6.90*10^14 Hz

b)

E(max) = B(max) * c, magnetic field amplitude, B(max) =

B(max) = E(max)/c

B(max) = 2.7*10^-3 / 3*10^8

B(max) = 9*10^-12 T

c)

1T = 1 (V.s/m^2)

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Explanation:

Given that,

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u=(1i+6j)\ m

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Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

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v=(1i+6j)+(5i+7j)10

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(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

y'=y+ut+\dfrac{1}{2}at^2

y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2

y'=\dfrac{19}{2}i+\dfrac{39}{2}j

(c) Magnitude of y',

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|y'| = 21.69 meters

Direction of the y',

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tan\theta=\dfrac{39/2}{19/2}

\theta=64.02^{\circ}

Hence, this is the required solution.

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3 years ago
I NEED HELP ASAP PLEASE!
REY [17]

Answer: thermal conductivity

7 0
3 years ago
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I appreciate any help I get on this, thank you in advanced :)
Reika [66]
The picture isn’t clear so I can’t read the dimensions of the box but I can try my best to guide u through the question.

For part a u need to find the volume of the box as that will equal the volume of sand that can be filled inside.
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For part b the mass of sand alone will be
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3 years ago
The watermark can be of different types depending on the application. A watermark will resist manipulations of the media. Howeve
Marina86 [1]

<em>The first blank is </em><em>robust watermark</em>; a robust watermark will not resist tampering.

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snow_lady [41]

Answer:

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The computation is shown below:

As we know that

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Torque = I\alpha

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