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iren [92.7K]
3 years ago
6

A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?​

Physics
1 answer:
Ilya [14]3 years ago
4 0

The frequency of the wave is 1\cdot 10^9 Hz

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

c=f \lambda

where

c is the speed of the wave

f is the frequency

\lambda is the wavelength

For the radio wave in this problem,

\lambda = 0.3 m

c=300,000,000 m/s = 3\cdot 10^8 m/s

Therefore, the frequency is:

f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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A transverse wave is set up in a very long string. The oscillator is set at 20.0 Hz, and the wave speed is 78 m/s. The amplitude
Sedbober [7]

To solve this problem we must apply the concepts related to Tangential Acceleration based on angular velocity and acceleration, and therefore, we must also calculate angular velocity based on the given frequency. For all these problems we will take the Units to the International System. The maximum acceleration would then be defined as,

a_{max} = \omega^2 A

Here,

\omega= Angular velocity

A = Amplitude

At the same time the angular velocity is described as,

\omega = 2\pi f

Here f means the frequency of the wave. Substituting,

\omega = 2 \pi (20)

\omega = 40\pi

A = 5.2cm

A = 0.052m

Replacing at the first equation,

a_{max} = (40\pi )^2 (0.052)

a_{max} = 821.15m/s^2

Therefore the maximum particle acceleration for a point on the string is 821.15m/s^2

6 0
3 years ago
What is the name of the process that is responsible for the formation of elements heavier than hydrogen? A) fusion B) fission C)
gladu [14]

Answer:

A) Fusion

Explanation:

Fusions of atoms heavier than hydrogen like helium takes place in the core of the star, at the core the temperature and density of gases is the greatest .

5 0
3 years ago
A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge
Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

E.A=\dfrac{q_{enclosed}}{\epsilon_o}

0=\dfrac{\lambda_i+\lambda}{\epsilon_o}

\lambda_i=-\lambda  

For outer surface,

\lambda_i+\lambda_o=2\lambda

-\lambda+\lambda_o=2\lambda

\lambda_o=3\lambda

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}

E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}

Hence, this is the required solution.

6 0
3 years ago
A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At
Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

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Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

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So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

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2 years ago
A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s.
Katarina [22]
Option B would be right one
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5400*3 = 16200kgm/s
16200-13200 = 3000
now 6600-5400 = 1200 kg
thus 3000/1200 = 2.5 v
5 0
3 years ago
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