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3 years ago

9

Where is visible light located on the electromagnetic spectrum?

2 answers:

Stells [14]3 years ago

7 0

Answer:

2. Near the middle

Explanation:

Visible light is a form of electromagnetic radiation which is visible to the human eye, it is located in the narrow band of wavelengths found on the left of the spectrum, with the ultraviolet wavelengths to the left and infrared to the right side of the spectrum. This places the visible light near the middle of the spectrum.

The human eyes have high sensitivity for detecting very narrow band of wavelengths despite the spectrum having a wide array of wavelengths. Visible light ranges from 400 nm to 700 nm

nikklg [1K]3 years ago

4 0

2 near the middle
between ultraviolet and infared

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A hummingbird flies forward and backward. Its motion is shown on the above graph of horizontal Position and

harina [27]

Answer:

The hummingbird travels the total distance is 10 m

Explanation:

Given that,

A hummingbird flies forward and backward.

According to figure,

We know that,

The total distance is equal to the sum of all distance.

Forward direction is positive and backward direction is negative.

We need to calculate the hummingbird travels the total distance

Using figure,

$D=d_{1}+d_{2}+d_{3}+d_{4}$

Put the value into the formula

$D=6+(-4)+2+6$

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Hence, The hummingbird travels the total distance is 10 m.

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4 years ago

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Nadya [2.5K]

Answer:

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3 years ago

Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl

Vera_Pavlovna [14]

Answer:

A, B, C, E

Explanation:

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3 years ago

Read 2 more answers

Boyle’s Law: When____ is held constant, the pressure and volume of a gas are___ proportional.

Aleksandr-060686 [28]

Answer:

When temperature is held constant, the pressure and volume of a gas are not proportional.

Explanation:

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3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

3 years ago