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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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kicyunya [14]
By definition, Ampere is a unit of current which is a measure of the amount of charge passing through a point in a circuit per unit  of time, with an equivalent charge of 1.602 x 10^(-19) Coulomb per electron. To determine the number of electrons passing through the heater, we use the definition of the current. We calculate as follows:

13.5 A = 13.5 C per second
Charge = 13.5 C/s (10 min) ( 60 s / 1 min)
Charge = 8100 C 

Number of electrons = 8100 C / 1.602 x 10^(-19) C per electron
Number of electrons = 5.1 x 10^22 electrons

Therefore, there are 5.1 x10^22 electrons that assed through the heater for 10 minutes.
3 0
4 years ago
How do your results from ray tracing compare to your results from using the thin-lens equation?
EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} where;

f is the focal length  of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm

Hence the focal length of the convex lens is 20cm

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3 years ago
Which of these changes will increase the energy loss of a building?
fenix001 [56]

1. The velocity decreases, and the kinetic energy decreases.

2. An increase in temperature difference between the inside and outside of the building.

3. The total kinetic energy remains the same.

4. 76,761 J

5. The energy loss must increase.

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3 years ago
A sealed box contains a monatomic ideal gas. The number of gas atoms per unit volume is 5.00 * 1020 atoms&gt;cm3, and the averag
Veronika [31]

Answer:

Pressure,P=6×10^3Pa

Explanation:

The gas has an ideal gas behaviour and ideal gas equation

PV=NKT

T= V/N p/K ...eq1

Average transitional kinetic energy Ktr=1.8×10-23J

Ktr=3/2KT

T=2/3Ktr/K....eq2

Equating eq1 and 2

V/N p/K = 2/3Ktr/K

Cancelling K on both sides

P= 2/3N/V( Ktr)

Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3

P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23

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3 years ago
A solar reflector is made using 31 identical triangular-shaped mirrors, each having sides . What is the total surface area of th
Stels [109]

Complete Question

Question 18 (3 points) Solve the problem. (3 points) A solar reflector is made using 31 identical triangular-shaped mirrors, each having sides 2.4m, 2. 3m, 1.5 m. What is the total surface area of the reflector?

A) 33 m2  

B) 86 m2

C) 52 m2

D)  34 m2

Answer:

The value is  Area  =2.78 \   m^2

Explanation:

From the question we are told that

   The  sides are  a =  2.4 m

                             b  =  2.3 m

                             c  =  1.5 m

Generally the semi perimeter is mathematically represented as

           s = \frac { a + b  + c  }{2 }

=>       s = \frac { 2.4  + 2.3  + 1.5   }{2 }

=>       s =3.1  \  m

Generally the using Heron's  formula we have that the  surface are a is  mathematically represented as

            Area  =  \sqrt{S (S -  a) (S - b )(S - c ) }

=>         Area  =  \sqrt{3.1  (3.1  -   2.4) (3.1  -  2.3 )(3.1  -  1.5 ) }

=>        Area  =2.78 \   m^2

6 0
3 years ago
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