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GarryVolchara [31]
3 years ago
9

A similarity between radio waves and microwaves is that both are used for _____.

Physics
2 answers:
Cloud [144]3 years ago
5 0

Answer:

(B) Sending Messages

Explanation:

  • A similarity between radio waves and microwaves is that both are used for sending messages and communication

Radio Waves and Microwaves have longer wavelengths and lower frequencies. Longer wavelengths than an x-ray which is in the electromagnetic spectrum and lower frequencies than gamma rays, which are also in the electromagnetic spectrum. They are also both electromagnetic radiation.

nikklg [1K]3 years ago
5 0
Sending messages because I said so and the guy on top said it so it’s right love u
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A technique in which people use machines to learn how to control their bodies is known as __________.
Debora [2.8K]

Answer:

The technique in which people use machines to learn how to control their bodies is known as D, Biofeedback.

Explanation:

Biofeedback is a variety of different machines that help people learn how to control their bodies depending on their specific needs, varying from things like scalp sensors, electrocardiographs, electromyographs and more.

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3 years ago
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How do organisms get the energy they need​
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consumers get energy from eating other things plant from sun

Explanation:

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4 years ago
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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

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

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
2. Use the diagram below to answer this question. As the ball moves from point A
Thepotemich [5.8K]

Answer:

at point A the ball possess pontetial energy , point B kinetic energy then point C pontetial energy

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If a bean of mass 2.0g jumps 1.0cm from your hand into the air, how much potential energy has it gained in reaching its highest
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PE = mg\Delta h = 0.002 \, kg \cdot 9.8 \, m/s^2 \cdot 0.01 \, m = ~2 \cdot 10^{-4} \, J
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