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2 years ago

What do radio waves and microwaves have in common?

Physics

2 answers:

barxatty [35]2 years ago

7 0

Answer:

Both are found next to visible light on the electromagnetic spectrum.

i know because i just took the test trust me i choose Both are found next to visible light on the electromagnetic spectrum. and got it wrong it said the answer was Both are found next to visible light on the electromagnetic spectrum.

Explanation:

forsale [732]2 years ago

3 0

Answer:

Both are at the side of the spectrum that has the lowest frequency.

Explanation:

We know that the electromagnetic spectrum is the set of several waves that are arranged in order of increasing wavelength of decreasing frequency. It consists of many waves like,

gamma rays, x- rays, ultraviolet rays, visible, infrared rays, microwave, radio waves.

Radio wave has the longest wavelength while gamma rays has maximum wavelength. From the attached figure, it is clear that the frequency of radio wave is least and that of microwave is second least frequency.

Hence, the correct option is (c).

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2 years ago

To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Nimfa-mama [501]

The energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
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$C=3.0 \mu F = 3.0 \cdot 10^{-6} F$
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
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3 years ago

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

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2 years ago