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finlep [7]
3 years ago
11

Electromagnetic waves are traveling in the vacuum of space. Calculate the wavelengths of these electromagnetic waves with the fo

llowing frequencies. (Enter the first wavelength in pm and the second wavelength in cm.)
(a) 2.00 x 1019 Hz
(b) 4.50 x109 Hz
Physics
1 answer:
Maksim231197 [3]3 years ago
4 0

Answer:

(a) 1.5×10⁻¹¹ m.

(b) 6.7×10⁻² m

Explanation:

Note: All Electromagnetic wave travels in with the same speed, which 3×10⁸ m/s

(a) Give a frequency of 2.00×10¹⁹ Hz.

Using the equation of a wave,

V = λf................ Equation 1

Where V = Speed of electromagnetic wave, λ = wavelength, f = frequency.

make λ the subject of the equation

λ = V/f................. Equation 2

Given: f = 2.00×10¹⁹ Hz.

Constant: v = 3×10⁸ m/s.

Substitute into equation 2

λ = 3×10⁸/2.00×10¹⁹

λ = 1.5×10⁻¹¹ m.

(b) Similarly using

λ = v/f

Given: f = 4.5×10⁹ Hz, and v = 3×10⁸ m/s.

Substitute these values into equation 2 above.

λ = 3×10⁸/4.5×10⁹

λ = 6.7×10⁻² m

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pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
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  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

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