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3 years ago

The wavelengths for visible light rays correspond to which of these options?

Physics

2 answers:

elena-s [515]3 years ago

6 0

Answer;

About the size of a virus or a large molecule

Explanation;

Visible light is a form of electromagnetic (EM) radiation, as are radio waves, infrared radiation, ultraviolet radiation, X-rays and microwaves. Generally, visible light is defined as the wavelengths that are visible to most human eyes.

Visible light falls in the range of the EM spectrum between infrared (IR) and ultraviolet (UV). It has frequencies of about 4 × 1014 to 8 × 1014 hertz (Hz) and wavelengths of about 740 nanometers (nm) to 380 nm .

The wavelength of visible light corresponds to the size of a virus or a large molecules which ranges from about 20 to 400 nanometres in diameter.

PilotLPTM [1.2K]3 years ago

6 0

The wavelengths for visible light rays correspond is about the size of a virus or a large molecule. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

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the amount of time for one particle of the medium to make one comlpete vibration cycle is a good description of

-Dominant- [34]

That's the definition of the PERIOD of the vibration.
It's exactly the reciprocal of the vibration's frequency.

7 0

3 years ago

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Question 4 (1 point)

Valentin [98]

cardiovascular endurance :))

4 0

3 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago

A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A

Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

Em₀ = U = mg y

Final. Low point just before the crash

Emf = K = ½ m v²

Em₀ = Emf

m g y = ½ m v²

v = √ 2 g y

Let's calculate

v = √ (2 9.8 0.05)

v = 0.99 m / s

1) the moment before the crash is

p₀ = m v

p₀ = 0.221 0.99

p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

p = 0

2) change of momentum

Δp = p - p₀

Δp = 0- 0.219

Δp = -0.219 kg m / s

3) the reason is

Δp / p = 1

In percentage form it is 100%

3 0

4 years ago

A crate falls from an airplane flying horizontally at an altitude of 2,000 m.

Hatshy [7]

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

7 0

4 years ago