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Arte-miy333 [17]
3 years ago
15

What does the wavelength of visible light waves correspond to?

Physics
2 answers:
marissa [1.9K]3 years ago
7 0

Explanation:

Visible light is a very part of electromagnetic spectrum. We can see the visible region of the electromagnetic spectrum.

The frequency and the wavelength has an inverse relationship.

For example, the red light has large wavelength and low frequency. Violet has a short wavelength and higher frequency.

The frequency range of visible region is from 400 nm to 700 nm.

Lesechka [4]3 years ago
7 0

The wavelength of the visible range of the electromagnetic spectrum lies between \boxed{400\,{\text{nm}} - 700\,{\text{nm}}} .

Further Explanation:

The Electromagnetic spectrum is the complete range of the electromagnetic radiation which expresses all the radiation starting from the Gamma rays to the radio waves. The electromagnetic spectrum is the detailed spectrum that explains all the electromagnetic radiations.

The electromagnetic radiations are the transverse waves which do not require any medium for their propagation. The electromagnetic waves can easily travel through space. All the electromagnetic radiations travel at the speed of light.

The different radiations in the electromagnetic spectrum are as follows:

  • Gamma rays
  • X-rays
  • Ultraviolet rays
  • Infrared rays
  • Micro waves
  • Radio waves

The wavelength and the frequency ranges for the different electromagnetic radiations is as shown in figure attached.

Therefore, the wavelength of the visible range of the light in the electromagnetic spectrum is \boxed{400\,{\text{nm}} - 700\,{\text{nm}}} .

Learn More:

1. What is the threshold frequency ν0 of cesium <u>brainly.com/question/6953278 </u>

2. Calculate the wavelength of an electron (m = 9.11 × 10-28 g) moving at 3.66 × 106 m/s <u>brainly.com/question/1979815 </u>

3. What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays <u>brainly.com/question/9059731 </u>

Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetic Waves

Keywords:

Electromagnetic spectrum, visible light, EM waves, wavelength, 700nm, 400 nm, 7000 A, frequency, gamma rays, transverse waves.

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The presence of a uniform magnetic field may be detected by using a
Alex777 [14]

Answer:

Magnetic compass

Explanation:

A magnetic field is a field that describes the magnetic effect of electric charges in a relative motion.

Magnetic field allows magnets to interact without contact. The principle involved in the function of the compass is that a magnetic field exerts a force on any moving charge and can be measured and detected by this effect.

8 0
3 years ago
Read 2 more answers
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Lemur [1.5K]

Answer:

a)   v = 0.9167 m / s,  b)  A = 0.350 m,  c)  v = 0.9167 m / s, d)  A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

          v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

           λ = x

           λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

          T / 2 = t

           T = 2t

period and frequency are related

           f = 1 / T

           f = 1 / 2t

we substitute

           v = x / 2t

           v = 5.50 / 2 3

           v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

          2A = y

           A = y / 2

           A = 0.700 / 2

           A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

      v = 0.9167 m / s

d) the amplitude is

           A = 0.500 / 2

           A = 0.250 m

4 0
3 years ago
Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi
insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

d

s−15

=

7

3

return trip is then :

d

s+15

=

7

5

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

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3 years ago
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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

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3 years ago
Which planet is to humankind?
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Earth. Only. Any other known planets are inapplicable.
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