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nalin [4]
3 years ago
5

A particular light source gives off light waves with a measured wavelength of

Physics
1 answer:
Umnica [9.8K]3 years ago
3 0

The frequency of the light source is 1.5 x 10¹⁵ Hz.

<h3>Frequency of the light source</h3>

The frequency of the light source is determined using the following equations;

c = fλ

where;

c is speed of light

f is the frequency

λ is the wavelength

f = (3 x 10⁸) / (2 x 10⁻⁷)

f = 1.5 x 10¹⁵ Hz

Thus, the frequency of the light source is 1.5 x 10¹⁵ Hz.

Learn more about frequency of light here: brainly.com/question/10728818

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PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is
IgorLugansk [536]

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, d_o = 15.0 cm

image distance, d_i = 5.0 cm

height of the object, h_o = 9.0 cm

height of the image, h_i = ?

Apply lens equation;

\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

4 0
3 years ago
What is the frequency of a wave having a period equal to 18 seconds <br>​
Ivanshal [37]

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

The relation between wave period and frequency is as follows.

T = \frac{1}{f}T=

f

1

where, T = time period

f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

T = \frac{1}{f}T=

f

1

or, f = \frac{1}{T}f=

T

1

= \frac{1}{18 sec}

18sec

1

= 0.055 per second (1cycle per second = 1 Hertz)

or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz

<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>
3 0
3 years ago
an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le
storchak [24]
<span>On what:

f (is the focal length of the lens) = ? 
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}
\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}
\frac{1}{f} =  \frac{10}{33.18}
Product of extremes equals product of means:
10*f = 1*33.18
10f = 33.18
f =  \frac{33.18}{10}
\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark

6 0
4 years ago
A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic
Veronika [31]

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

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current  i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28  × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

3 0
3 years ago
What is the difference between 5 mL of water and 5.0 mL of water?
ad-work [718]
There is no difference. 5 is the same as 5.0
6 0
4 years ago
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