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Varvara68 [4.7K]
3 years ago
6

Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is

most sensitive to light having a wavelength of 5.50 Light waves are electromagnetic waves that travel 10-7 m.(a) Find the frequency of this light wave.Hz(b) Find its period.s
Physics
1 answer:
svlad2 [7]3 years ago
5 0

(a) 5.45 \cdot 10^{14} Hz

The relationship between frequency and wavelength of an electromagnetic wave is given by

c=f \lambda

where

c=3.00 \cdot 10^8 m/s is the speed of light

f is the frequency

\lambda is the wavelength

In this problem, we are considering light with wavelength of

\lambda=5.50 \cdot 10^{-7} m

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz

(b) 1.83\cdot 10^{-15} s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

And using f=5.45 \cdot 10^{14} Hz as we found in the previous part, we can find the period of this wave:

T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s

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A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-
musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0
4 years ago
A star's emission spectrum consists of
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A star's emission spectrum consists of color lines indicating wavelength of light, black lines indicating wavelength of light, and a rainbow of all the colors of visible light; one red light.
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4 years ago
A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa
Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because    v_1 =v_2 =v_3

b

 The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

                                w_1 >w_2 = w_3  

c

The ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

                          r_2 =r_3 >r_1

Explanation:

Mathematically Frequency can be represented as

                         F = \frac{v}{\lambda}

Where \lambda is the wavelength and v is the velocity

   Now looking at the diagram we see that

          For the  first frequency we have

             Let the wavelength be  \lambda_1 = \lambda , and the frequency  F_1 = F

           For  the second frequency

           Let the wavelength be  \lambda_2 = 2 \lambda , and the frequency F_2 = \frac{F}{2}

           For  the third frequency

           Let the wavelength be  \lambda_3 = 2\lambda ,  and the frequency F_3 = \frac{F}{2}

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

  So,

        For the  first frequency we have

                                 v_1 = \lambda_1 F_1 = \lambda F

          For  the second frequency

                               v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F      

           For  the third frequency

                               v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F

Hence

The Ranking of the curve according to their speed would be equal Rank because    v_1 =v_2 =v_3

 Mathematically angular speed can be represented as

                           w = 2 \pi f

   For the  first frequency we have

                          w_1 = 2\pi F_1 = 2 \pi F                        

    For  the second frequency

                        w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2}  = \pi F

     For  the third frequency

                      w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2}  = \pi F  

 Hence

          The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

                                w_1 >w_2 = w_3  

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

                            v = wr

                    =>    r = \frac{v}{w}

 Since the linear velocity is constant we have that

                            r \  \alpha \  \frac{1}{w}

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of  the second third frequency would be the same but their ranking would be greater than that of the first frequency because

                          r_2 =r_3 >r_1

       

5 0
3 years ago
In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm,
Serga [27]

Answer:

4E

Explanation:

From the question given above, the following data were obtained:

Initial elongation (e₁) = 4 cm = 4/100 = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.04 + 0.04 = 0.08 m

Final energy (E₂) =?

The energy stored in a s spring is given by:

E = ½Ke²

Where

E => is the energy

K => is the spring constant

e => is the elongation

From:

E = ½Ke²

Energy is directly proportional to the elongation. Thus,

E₁/e₁² = E₂/e₂²

With the above formula, we can obtain the final energy as follow:

Initial elongation (e₁) = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.08 m

Final energy (E₂) =?

E₁/e₁² = E₂/e₂²

E / 0.04² = E₂ / 0.08²

E / 0.0016 = E₂ / 0.0064

Cross multiply

0.0016 × E₂ = 0.0064E

Divide both side by 0.0016

E₂ = 0.0064E / 0.0016

E₂ = 4E

Therefore, the final energy is 4 times the initial energy i.e 4E

6 0
3 years ago
a banana boat accelerates from 4.167 m/s ar 2.00m/s^2.How far has it traveled when it reaches 8.333 m/s
Nataly_w [17]

Answer: The distance that we'll be travelled is d=16.35m

Explanation: The main idea here is to use the equation of motion. We are given the acceleration of 2m/s^2,the initial and final velocies. The formula to use is v^2=u^2+2as. Now we have to substitute the values.

2^2= 8.33^2+2(2)d

d= 16.35m

8 0
4 years ago
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