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Helen [10]
3 years ago
9

What determines the frequency of an electromagnetic wave

Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

<h2>The frequency depends upon the source energy of electromagnetic waves</h2>

Explanation:

The energy of electromagnetic  wave is

E  = h ν

E is the energy of source emitting the waves

here h is Plank's constant

and ν is used for the frequency of the electromagnetic wave

Thus the frequency of wave depends upon the energy of electromagnetic source . Because h is constant .

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U WILL GET 25 POINTS PLEASE HELP
balandron [24]

Answer:

Hopefully, this is right.

Explanation:

Answer is 5.43.

3 0
3 years ago
An object has a horizontal velocity component of 6 m/s and a vertical velocity component of 4 m/s.
taurus [48]

For the object that has a horizontal velocity component of 6 m/s and a vertical velocity component of 4 m/s we have:

1. The velocity of the object is 7.21 m/s.

2. The angle it makes with the horizontal is 33.7°.

1. The velocity of the object can be found as follows:

v = \sqrt{v_{x}^{2} + v_{y}^{2}}

Where:

v_{x}: is the horizontal component of the velocity = 6 m/s

v_{y}: is the vertical component of the velocity = 4 m/s

Hence, the <u>velocity is</u>:

v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(6 m/s)^{2} + (4 m/s)^{2}} = 7.21 m/s

2. The angle it makes with the <u>horizontal </u>can be calculated with the following trigonometric function:

tan(\theta) = \frac{v_{y}}{v_{x}}

Where:

θ: is the angle it makes with the horizontal

Therefore, the <u>angle is</u>:

\theta = tan^{-1}(\frac{v_{y}}{v_{x}}) = tan^{-1}(\frac{4}{6}) = 33.7

You can learn more about the components of the velocity here: brainly.com/question/2285233?referrer=searchResults

I hope it helps you!

7 0
3 years ago
During the Patient Care Era, a more demanding and educated public now expected the pharmacist to be a therapeutic advisor and __
Murrr4er [49]
Make a lot of money for what they do
3 0
4 years ago
What is the relationship between chromosomes and a body's appearance? (A) Chromosomes make it possible for all bodies to exactly
Katen [24]

Answer: The correct option is B.

Chromosomes contain genes that tell the body how to grow and work.

Explanation:

This is because chromosomes contain alot of human genes which are found in the nucleus and every humans have 23 pairs of chromosomes. The sets of genes determines some of the body features, characteristics or traits. The genes in the body form the genotype and the genotype is expressed physically which is now the phenotype i.e traits expressed physically in the body which is an indication of body appearance.

7 0
3 years ago
At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r
Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s    

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.                

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

W = \Delta Vq    (1)                                        

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C    

V: is the potential    

Also, the work is equal to:

W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}     (2)      

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

v_{a}: is the velocity in the point a

v_{b}: is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

\Delta Vq = \frac{1}{2}mv_{a}^{2}

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}

v_{a} = 2.9 \cdot 10^{4} m/s

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}

Since v_{a} = 0 (starts from rest) and m_{e} = 9.1x10⁻³¹ kg (electron's mass), we have:

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}    

v_{b} = 1.2 \cdot 10^{6} m/s

I hope it helps you!

6 0
4 years ago
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