What is the wavelength of an ekectromagnetic wave that has a frequency of 40,000 hz? the speed of light is 3x10^8 m/s?

A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i

nexus9112 [7]

Answer:

E = k q / a² (1.3535) (- i ^ + j ^)

E = k q / a² 1.914 , θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

E₁₂ = k q / a²

On the y axis

E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

d = √ (a² + a²) = a √2

let's look for the field

E₁₃ = k q / d²

E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

cos 45 = E₁₃ₓ / E₁₃

E₁₃ₓ = E₁₃ cos 45

sin 45 = E_{13y} / E₁₃

E_{13y} = E₁₃ sin 45

E₁₃ₓ = k q / 2a² cos 45

E_{13y} = k q / 2a² sin 45

let's find each component of the electric field

X axis

Eₓ = -E₁₂ - E₁₃ₓ

Eₓ = - k q / a² - k q / 2a² cos 45

Eₓ = - k q / a² (1 + cos 45/2)

cos 45 = sin 45 = 0.707

Eₓ = - k q / a² (1 + 0.707 / 2)

Eₓ = - k q / a² (1.3535)

Y axis

E_y = E₁₄ + E_{13y}

E_y = k q / a² + k q / 2a² sin 45

E_y = k q / a² (1 + sin 45/2)

E_y = k q / a² (1.3535)

we can give the results in two ways

E = k q / a² (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

E = √ (Eₓ² + E_y²)

E = k q / a² 1.3535 √2

E = k q / a² 1.914

we use trigonometry for the angle

tan θ = E_y / Eₓ

θ = tan⁻¹ (E_y / Eₓ)

θ = tan⁻¹ (1 / -1)

θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

θ‘= 90 + 45

θ’= 135

4 0

3 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

2 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

1 year ago

A bird lands on a bare copper wire carrying a current of 51

nirvana33 [79]

The resistance of the piece of wire is
$R= \frac{\rho L}{A}$
where
$\rho = 1.68 \cdot 10^{-8}\Omega m$ is the resistivity of the copper
$L=5.1 cm=0.051 m$ is the length of the piece of wire
$A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2$ is the cross sectional area of the wire
By substituting these values, we find the value of R:
$R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega$

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
$V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V$

7 0

3 years ago

To prevent accidental electric shock, most electrical equipment is A. painted. B. isolated. C. coated. D. grounded.

Nesterboy [21]

D. It is Grounded below ground

4 0

3 years ago

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