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barxatty [35]
3 years ago
15

Your local AM radio station broadcasts at a frequency of f = 1100 kHz. The electric-field component of the signal you receive at

your home has the time dependence E(t) = E0 sin(2πft), where the amplitude is E0 = 0.62 N/C. Radio waves travel through air at approximately the speed of light.
a) At what wavelength, in meters, docs this station broadcast?
b) What is the value of the radio wave's electric field, in newtons per coulomb, at your home at a time of t = 3.1 μs?
Physics
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

(a) 272.73 m

(b) 0.338 N/C

Explanation:

frequency, f = 1100 kHz = 1100 x 1000 Hz

E(t) = Eo Sin(2πft)

Eo = 0.62 N/C

(a) Velocity of light, c = 3 x 10^8 m/s

wavelength, λ = c / f = (3 x 10^8) / (1100000) = 272.73 m

Thus, the wavelength is 272.73 m.

(b) at t = 3.1 microsecond = 3.1 x 10^-6 s

E = Eo Sin (2 π ft)

E = 0.62 Sin (2 x 3.14 x 1100 x 10^3 x 3.1 x 10^-6)

E = 0.62 Sin (21.4148)

E = 0.62 x 0.5449 = 0.338 N/C

Thus, the electric field at t = 3.1 microsecond s 0.338 N/C.

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

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