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satela [25.4K]
3 years ago
15

How do you suppose the frequency of an electromagnetic wave compares with the frequency of the electrons it sets into oscillatio

n in a receiving antenna?
Physics
1 answer:
FromTheMoon [43]3 years ago
6 0

Answer:

The frequencies are the same.

Explanation:

This is similar to the photoelectric effect. The frequency of the electromagnetic wave has to match a threshold frequency before it can set electrons into oscillation in a receiving antenna. After the frequency of the electromagnetic waves have matched and exceeded this threshold frequency, all of it's energy is converted to the kinetic energy of the electrons it sets into oscillation. And since their energies are similar, the frequencies too, subsequently, will be the same.

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Pls answer quick I need to get the answer rn
lora16 [44]
I think it is False because as the Gad relajases fuel it doesn’t move as much anymore
3 0
2 years ago
(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

6 0
3 years ago
For a concave mirror, when the object is at focus the image is formed at
Sergeeva-Olga [200]

Answer:

When the object is placed at the focus the image is formed at infinity.

Explanation:

When a ray passes through focus and incident on a concave mirror then it will travel parallel to principal axis after reflection.Hence the image is formed at infinity.

8 0
3 years ago
Assume that the function x(t) represents the length of tape that has unwound as a function of time. find θ(t), the angle through
bekas [8.4K]
We know that arc length (x(t)) is given with the following formula:
x(t)=\theta(t) r
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
r(\theta)=r_0-\frac{D\cdot \theta}{2\pi}
When we plug this back into the first equation we get:
x(t)=\theta(r_0-\frac{D\theta}{2\pi}})\\ \frac{D\theta^2}{2\pi}-\theta r_0+x(t)=0\\
We must solve this quadratic equation.
The final solution is:
\theta=\frac{\pi r_0+\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D},\:\theta=\frac{\pi r_0-\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D}
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
x(t)=\theta(r_0-\frac{D\theta}{2\pi}});D=0\\
x(t)=\theta r_0\\
\theta(t)=\frac{x(t)}{r_0}

8 0
3 years ago
Express 50,000 m/s2 in scientific notation. 5.0 x 103m/s2 0.5 x 105m/s2 5.0 x 104m/s2
Tresset [83]
Answer would be 5.0x104m/s2
8 0
3 years ago
Read 2 more answers
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