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2 years ago

Hazel recently read an article that claimed that radio waves can cause electron displacement. Based on

Physics

1 answer:

iren [92.7K]2 years ago

3 0

Answer:

Based on the passage, the article's claim is not true. Therefore, the answer is:

No, because radio waves have energies that are too low to fall in the ionizing range.

Explanation:

Radio waves are generally known as forms of non-ionizing radiation. This means that they do not have enough energy, unlike microwaves, to separate electrons from atoms or molecules, thereby ionizing them. They cannot cause electron displacement, as claimed in the article. Therefore, they cannot break chemical bonds, which can cause chemical reactions or DNA damage. As non-ionizing radiation, radio waves occur at lower frequencies in the electromagnetic spectrum.

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Which of the following will be the best condition for bread fermentation?

dezoksy [38]

Answer:

a. Wet, soft dough at 85 degrees Fahrenheit

Explanation:

Fermentation is an anaerobic process that transforms starches into simpler substances. The rising of dough is due to fermentation.

According to Harold McGee, 85°F (29°C) is the best temperature for fermenting bread dough. Temperatures below 85°F (29°C) take much longer to ferment, and temperatures higher than that result into unpleasant flavors in the dough.

Wet, soft dough is usually more preferable because it produces a softer bread.

3 0

3 years ago

What type of wave does not need matter to carry energy?

marishachu [46]

Non- mechanical wave does not need matter to carry energy.

e.g:- Light

3 0

3 years ago

Read 2 more answers

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the pow

allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

$l$ = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

$l$ = $l$ 1 + $l$ 2 + 2√ $l$ 1 $l$ 2cos(∅)

where

$l$ 1 = 2.25 mW

$l$ 2 = 2.025 mW

∅ = π

so we substitute

$l$ = 2.25 + 2.025 - 2√(2.25 × 2.025)

$l$ = 4.275 - 4.26907

$l$ = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

5 0

2 years ago

describe an experiment to show how the frequency of a note emitted by a vibrating string depends on the tension of the string

mart [117]

Easy !

Take any musical instrument with strings ... a violin, a guitar, etc.

The length of the vibrating part of the strings doesn't change ...
it's the distance from the 'bridge' to the 'nut'.

Pluck any string. Then, slightly twist the tuning peg for that string,
and pluck the string again.

Twisting the peg only changed the string's tension; the length
couldn't change.

-- If you twisted the peg in the direction that made the string slightly
tighter, then your second pluck had a higher pitch than your first one.

-- If you twisted the peg in the direction that made the string slightly
looser, then your second pluck had a lower pitch than the first one.

3 0

3 years ago

A 12.0-g bullet is fi red horizontally into a 100-g wooden block initially at rest on a horizontal surface. After impact, the bl

allochka39001 [22]

Answer:

$v_1 = 91.3 m/s$