PHYSICS COLLEGE

A radio station broadcasts its electromagnetic (radio)waves at a frequency of 9.05 x 107 Hz.
These radio waves travel at a speed of 3.00 x 108 m/s.
What is the wavelength of these radio waves?

Answers

Answer 1
Answer: Wavelength = speed/frequency
Wavelength = 3.00x108/9.05x107=
3.3x10 risen to the power of -1

Related Questions

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
In a region where there is a uniform electric field that is upward and has magnitude 3.80x104 N/C a small object is projected upward with an initial speed of 2.32 m/s The object travels upward a distance of 5.98 cm in 0 200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)?Assume g 9.80 m/s and ignore air resistance E3? C/kg q/m
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A rectangular loop (area = 0.15 m2) turns in a uniform magnetic field, B = 0.18 T. When the angle between the field and the normal to the plane of the loop is π/2 rad and increasing at 0.75 rad/s, what emf is induced in the loop?
Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit. Required:
Find the radius and period of the orbit.

Answers

Answer:

  r = 2,026 10⁹ m  and   T = 2.027 10⁴ s

Explanation:

For this exercise let's use Newton's second law

        F = m a

where the force is electric

        F = k (q_1q_2)/(r^2)

Acceleration is centripetal

        a = v² / r

we substitute

        k (q_1q_2)/(r^2) = m (v^2)/(r)

        r = k (q_1q_2)/(m \ v^2)          (1)

let's look for the charge in the insulating sphere

          ρ = q₂ / V

          q₂ = ρ V

the volume of the sphere is

         v = 4/3 π r³

we substitute

        q₂ = ρ (4)/(3) π r³

        q₂ = 3 10⁻⁹ (4)/(3) π 4³

        q₂ = 8.04 10⁻⁷ C

let's calculate the radius with equation 1

        r = 9 10⁹  1.6 10⁻¹⁹  8.04 10⁻⁷ /(9.1  10⁻³¹ 628 10³)

        r = 2,026 10⁹ m

this is the radius of the electron orbit around the charged sphere.

Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio

        v = x / t

the distance traveled in a circle is

        x = 2π r

In this case, time is the period

        v = 2π r /T

        T = 2π r /v

let's calculate

        T = 2π 2,026 10⁹/628 103

        T = 2.027 10⁴ s

The height of the upper falls at Yellowstone Falls is 33 m. When the water reaches the bottom of the falls, its speed is 26 m/s. Neglecting air resistance, what is the speed of the water at the top of the falls?

Answers

Answer:

Speed of water at the top of fall = 5.40 m/s

Explanation:

We have equation of motion

v^2=u^2+2as

Here final velocity, v = 26 m/s

a = acceleration due to gravity

a=9.8m/s^2 \n

displacement, s = 33 m

Substituting

26^2=u^2+2* 9.8 * 33\n\nu^2=29.2\n\nu=5.40m/s \n

Speed of water at the top of fall = 5.40 m/s

Which of the following wouldhave low electromagnetic
energy?
A. X-rays
B. ultraviolet waves
C. radio waves

Answers

Final answer:

Radio waves have low electromagnetic energy compared to X-rays and ultraviolet waves.


Explanation:

Electromagnetic energy refers to the energy associated with electromagnetic waves, which are a form of energy that can travel through empty space. The energy of an electromagnetic wave is directly proportional to its frequency. Therefore, the frequency determines the energy level of the wave.

In the given options, radio waves would have the lowest electromagnetic energy. Radio waves have the longest wavelength and lowest frequency among the three options. X-rays, on the other hand, have a higher frequency and shorter wavelength, making them more energetic. Ultraviolet waves have an even higher frequency and shorter wavelength, making them the most energetic among the three options.


Learn more about Electromagnetic energy here:

brainly.com/question/22963020

Each plate of a parallel‑plate capacitor is a square of side 0.0479 m, and the plates are separated by 0.479 × 10 − 3 m. The capacitor is charged and stores 8.11 × 10 − 9 J of energy. Find the electric field strength E inside the capacitor.

Answers

Explanation:

It is known that the relation between electric field and potential is as follows.

             E = (V)/(d)

And, formula to calculate the capacitance is as follows.

           C = (\epsilon_(o) A)/(d)

              = (8.854 * 10^(-12) * (0.479 m)^(2))/(0.479 * 10^(-3))

              = 4.24 * 10^(-9) F

Hence, energy stored in a capacitor is as follows.

         W = (1)/(2)CV^(2)

          V = \sqrt{(2W)/(C)}

        E = \sqrt{(2W)/(d^(2)C)}

            = (2 * 8.11 * 10^(-9) J)/((0.479 * 10^(-3))^(2) * 4.24 * 10^(-9))

            = 16.687 * 10^(3) N/C

Thus, we can conclude that electric field strength E inside the capacitor is 16.687 * 10^(3) N/C.

. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force

Answers

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration  

              ∑ F = m a

in this case there are two forces on the x axis

             F_applied - fr = 0

since they indicate that the velocity is constant, consequently

             F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.

How fast can the car take this curve this curve without skidding to the outside of the curve?

Answers

Lets write the data down. That will help us solve the problem later:

R = 36 m

θ = 18º

m = 1492 kg

μ = 0.67

g = 9.8 m/s²

Lets draw all the forces that act on the car:

In order to the car won't skidding to the outside of the curve we must have the centripetal force equals the friction force:

F_(cp)=f_a

(m.v^(2))/(R)=\mu.F_N
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