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3 years ago

The highest frequency radio waves are which type ?

Physics

1 answer:

Zolol [24]3 years ago

7 0

Answer: Gamma rays

Explanation: Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies.

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

GaryK [48]

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

$y = \frac{m\lambda}{d}$

Here,

m = Number of order bright fringe

$\lambda$ = Wavelength

d = Distance between slits

Both distance are the same, then

$y_1 = y_2$

$\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}$

$\frac{m_1\lambda_1}{m_2\lambda_2} =1$

Rearranging to find the second wavelength

$m_1 \lambda_1 = m_2 \lambda _2$

$\lambda_2 = \frac{m_1\lambda_1}{m_2}$

$\lambda_2 = \frac{7(629)}{8}$

$\lambda_2 = 550.3nm$

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

7 0

3 years ago

Does Anybody Know The Answers?

ch4aika [34]

Answer:

I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry

Explanation:

$9^{2} + 12^2 = x^2\\81 + 144= x^2\\\sqrt{225} = \sqrt{x} \\ 15=x\\\\ 2.\\x^2+12^2+=13^2\\x^2+144 =169\\x^2 = 25\\\sqrt{x^2 =\sqrt{25\\\\$

x=5

$3.\\12^2+32^2 = x^2\\34.176= x$

5,12,13

$\frac{x}{4} ,\frac{12}{4} ,\frac{20}{4}\\\\\frac{x}{4},3,5 \\\\x=16\\\\12. \\x^2 + 48^2=50^2\\\\x^2=196\\x=14$

Download docx

5 0

3 years ago

Where do the electrons that form the auroras enter the magnetosphere? a. Through holes c. At the equator b. Between the magnetic

marta [7]

I think the answer is d. In the magnetotail. I hope this helps! :)

7 0

4 years ago

Read 2 more answers

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

A rocket is fired at 100 m/s at an angle of 37, what was its speed at the top of its path?

Nadusha1986 [10]

Answer:

0m/s

Explanation:

Since its fired at an angle, at the top there will be a split second where the velocity will be 0, as it has a parabolic shape, so the speed at the top of its path is 0

4 0

3 years ago